### arvindf232's blog

By arvindf232, 22 months ago,

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## 1734A — Select Three Sticks

We first sort the array $a$ in non-decreasing order.

Denote the indices of the elements that we choose from $a$ to be $x$, $y$, and $z$, where $1 \le x < y < z \le n$, and the final value (after performing the operations) of the concerned elements to be $v$.

The minimum required number of operations is then $|a_x-v|+|a_y-v|+|a_z-v|$. It is well-known that such expression attains its minimum value when $v$ is the median of $a_x$, $a_y$, and $a_z$. Since the array $a$ has already been sorted, it is best to assign $v$ to be $a_y$.

Our expression then becomes $|a_x-a_y|+|a_y-a_y|+|a_z-a_y|=(a_y-a_x)+0+(a_z-a_y)=a_z-a_x$. We would like to minimize the value of $a_z$, which implies $z$ should be as small as possible since $a$ is sorted. It is clear that taking $z=y+1$ would minimize the value of the expression. Similarly, we can show that we can take $x=y-1$ to minimize the value of the expression.

Therefore, the only possible values of the triplets $(x,y,z)$ are of the form $(t,t+1,t+2)$ for positive integers $1 \le t \le n-2$, and we can iterate through all such triplets and find the best one.

The time complexity is $\Theta(n \log n)$ per case due to sorting.

Code in C++

## 1734B — Bright, Nice, Brilliant

Note that the brightnesses of the rooms on the $i$-th floor is at most $i$. This is because in room $(i,1)$, only $i$ rooms, namely, $(1,1)$, $(2,1)$, $\ldots$, $(i,1)$ can reach to $(i,1)$ through some number of staircases.

It is also possible to find a configuration of torches in the pyramid such that the brightnesses of the rooms on the $i$-th floor is exactly $i$, i.e. it attains the upper bound.

The configuration is as follows: Room $(i,j)$ contains a torch if and only if it is the leftmost room ($i=1$) or the rightmost room ($i=j$) on the $i$-th floor.

This is valid because for all rooms $(i,j)$, it can be reached from $(1,1)$, $(2,1)$, $(3,1)$, $\ldots$, $(i-j+1,1)$ and $(2,2)$, $(3,3)$, $\ldots$, $(j,j)$. In other words, room $(i,j)$ has brightness $(i-j+1)+j-1=i$, so the pyramid is nice.

Code in C++

## 1734C — Removing Smallest Multiples

One operation should be used to remove every element not belonging to $T$.

Let $v$ be an element not belonging to $T$. Suppose a $x$-cost operation removes value $v$, then $v$ must be divisible by $x$. Furthermore, the multiples $x,2x,\cdots (k-1)x$ must have been already removed from $S$, where we write $v = kx$.

Since removed elements stay removed, the above is only possible if all of $x,2x,\cdots (k-1)x$ does not belong to $T$.

For each $v$, let $f(v)$ be the smallest integer $x$ satisfying the above condition. As we can always remove $v$ using a $v$-cost operation, $f(v) \leq v$ and in particular $f(v)$ exists.

The total cost must be at least $\sum_{i \not \in T} f(i)$. We claim that this cost can be achieved.

To do so, we should remove the required elements in ascending order. When removing $v$, we assume all $w \not\in T$ with $w<v$ have already been removed. At this state, an $f(v)$-cost operation would be able to remove $v$.

It remains to find the values $f(v)$. To do so efficiently, we can perform the above process in a bottom-up manner similar to the Sieve of Eratosthenes. Please refer to the code below for implementation details. The overall complexity is $n (1 +\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}) = \Theta(n \log n)$.

Code in C++

## 1734D — Slime Escape

Let's call a group of slime good if their total health is at least $0$, or if defeating this group allows you to reach the exits.

We partition the slimes into good groups in a two-pointer like manner. To form the groups to the right, start from position $k$, then find the smallest position $r$ such that slimes from $k+1$ through $r$ form a good group. We do the same starting from $r+1$ again. Repeat this process until slimes to the right are partitioned into groups, which can be done by maintaining the sum of health. We partition the left slimes into groups in a similar way.

We can observe that in an optimal strategy, we may assume the player absorbs group-by-group.

Proof

For any good group, since the total health is positive, there is no drawback to absorbing a good group. In other words, whenever it is possible to absorb a good group, we will absorb it.

For each group $G$, we calculate the requirement'' of the group — the lowest health we can begin with, such that we can absorb the group while maintaining non-negative health at all times. The requirement of a group of slime with health $a_1,a_2 \cdots a_n$ can be expressed as

$- \min_{k=0}^n (\sum_{i=1}^k a_i)$

Finally, we can simply simulate the process. We repeatedly attempt to absorb good groups to the left or to the right. We keep track of the current health, initially equal to $a_k$. Whenever we consider whether to absorb a group or not, we absorb it if and only if the current health is at least as high as its requirement. Otherwise, we ignore it for now and attempt to do so for the group on the other side. If it is possible to reach a state where either all left groups or all right groups are absorbed, then we can win the game. If at some point, it is not possible to absorb the left group nor the right group, then we lose.

The overall complexity is $\Theta(n)$.

It is also possible to use a range $\max$/$\min$ segment tree form the groups instead of using two-pointers, in which case its complexity would be $\Theta(n \log n)$.

Code in C++

## 1734E — Rectangular Congruences

We say a matrix to be good if it satisfies the congruence condition (the second condition).

When we have a good matrix, we can add any value $c$ to a whole row while maintaining the congruence relation. The same is true for adding the same value to a whole column.

Suppose we have any good matrix $A$, then by adding $b_i - a_{i,i}$ to the $i$-th row for each of $i=1,2,\cdots,n$, we obtain a good matrix that has the desired values on the diagonal.

In fact, there are a lot of possible constructions. We present a few of them here:

• $a_{i,j} = i \times j \pmod n$

• $a_{i,j} = (i + j)^2 \pmod n$. This needs special handling when $n=2$.

• $a_{i,j} = \frac{(i+j)(i+j+1)}{2} \pmod n$.

• The coolest part is that all quadratic polynomials in the form $ai^2 + bij + cj^2 + di + ej + f$ are valid for all integers $a,b,c,d,e,f$ and $b \not\equiv 0 \pmod n$

As a bonus, we prove that the general quadratic polynomial gives a good construction.

Proof

Here are some extra observations that may enable one to find a good matrix more quickly.

Some extra observations
Code in C++

## 1734F — Zeros and Ones

Observe that the $i$-th character is 1' if and only if $i$ has an odd number of set bits in its binary representation. Both solutions make use of this fact.

The constraints allows solutions of up to $\Theta(q \log^3 n)$. Yet, both of the model solution runs in $\Theta(\log n)$.

### 1. Digit DP solution

The question can be reformulated as follows: How many integers $x$ between $0$ and $m-1$ inclusive have the property that the total number of set bits of $x$ and $x+n$ is an odd number?

This can be solved with digit DP. We process the bit position from $\lceil \log(\max) \rceil$ down to $0$. We maintain three states:

• $\text{ans}$, a boolean value;

• $\text{trailzeros}$, an integer between $0$ and $\lceil \log(\max) \rceil$ inclusive; and

• $\text{under}$, a boolean value.

We can thus conclude the following: the number of trailing zeros is all we need to decide the answer.

After processing each bit $k$, we should have the following: the number of integers $x$ between $0$ and $\lfloor \frac{m}{2^k} \rfloor$ inclusive which have the following property:

• the total number of set bits of $x$ and $x + \lfloor \frac{n}{2^k} \rfloor$ is equal to $\text{ans} \mod 2$;

• the number of trailing 1's of $x + \lfloor \frac{n}{2^k} \rfloor$ is equal to $\text{trailzeros}$;

• the boolean value $[x < \lfloor \frac{m}{2^k} \rfloor]$ (where $[]$ is the Iverson bracket).

Now onto the transitions. Suppose we are adding the $(k-1)$-th digit, and let $d$ be the new digit of $x$, and $z$ be the $(k-1)$-th digit of $n$.

• If $z+d = 0$, then $(\text{ans},\text{trailzeros})$ after digit $k$ will be transited to $(\text{ans},0)$ after digit $k-1$;

• if $z+d = 1$, then $(\text{ans},\text{trailzeros})$ after digit $k$ will be transited to $((\text{ans} + z +d) \mod 2, \text{trailzeros}+1)$ after digit $k-1$;

• if $z+d =2$, then $(\text{ans},\text{trailzeros})$ after digit $k$ will be transited to $((\text{ans} + z + \text{trail} + 1) \mod 2, 0)$ after digit $k-1$

The final answer is the total number of values for which $\text{ans} =1$ and $\text{under} = 1$.

The above solution runs in $\Theta(\log^2 (\max))$ per query. There is a simple way to optimize this to $\Theta(\log(\max))$: note that we only need to keep parity of $\text{trailzero}$.

There are many other digit DP approaches that give similar time complexity. The constraints should allow most of them pass.

Code in Kotlin

### 2. Recursive Solution, by darkkcyan

Define the function $f(x) :=$ the parity of bit one in the number $x$.

We have thus reworded the statement into evaluating the follow expression:

$T = \sum_{i = 0}^{k - 1} \left[f(i) \ne f(i + n)\right]$

The formula can be further transformed as:

$T = \sum_{i = 0}^{k - 1} f(i \oplus (i + n))$

since $\left[ f(a) \ne f(b) \right] = f(a \oplus b)$ holds true for all non-negative integers $a$ and $b$.

Imagine we construct a grid and assign the value at row $r$ and column $c$ to be $f(r \oplus c)$. Then, $T$ is sum of a diagonal of length $k$ which starts at either $(0, n)$ or $(n, 0)$. Without loss of generality, we use $(0, n)$ in this editorial.

The grid can be constructed similarly to the way we construct the string $S$. We start with a $1$-by-$1$ matrix $M_0=\begin{bmatrix} 0 \end{bmatrix}$.

Then, the matrix $M_i$ of size $2^i \times 2^i$ can be constructed as follows:

$M_i = \begin{bmatrix} M_{i - 1} & \overline {M_{i - 1}} \\ \overline{M_{i - 1}} & M_{i - 1} \end{bmatrix}$

where $\overline{M_{i - 1}}$ is the matrix $M_{i - 1}$ but with flipped bits.

Here is another way of constructing the grid: let $C_i$ be an infinite chess board with alternating colors, similar to a regular chess board, but with each of the cells being size $2^i \times 2^i$. For example, $C_0$, $C_1$ and $C_2$ in an $8$-by-$8$ grid is as follows:

Illustration

We claim that our grid is the $\text{xor}$ of all chess board $C_i$. The proof is easy: $C_i$ is constructed by $\text{xor}$-ing the $i$-th bit of the the row and column number.

We are therefore motivated to proceed in the following way: if we drop the least significant bit (by making it to be $0$), we are still solving a very similar problem to the original problem, because dropping the first bit is similar to removing $C_0$. And when we shift $C_i$ to $C_{i - 1}$, it is a recursion of the same problem!

Going back to the problem, where we are computing sum of a diagonal of length $k$. If $k$ is odd, we can make it odd by adding the last element to the result and decreasing $k$ by one. Now, $k$ is even, and we can utilize the recurrence as follows:

• remove $C_0$.

• scale the board down by 2 (including $n$ and $k$). By doing so, $C_i$ becomes $C_{i - 1}$.

• solve the new problem.

• scale the board up again and add $C_0$ back.

• from the result of the scaled down problem, some how calculate the result of the original problem

The result of the scaled down problem is the number of $2$-by-$2$ cells with value $1$. From the number of $2$-by-$2$ cells with value $1$, we compute the number of cells with value $0$ as well. It is not hard to observe that it crosses the $2$-by-$2$ cells at all places. The only thing that matters is the parity of $n$.

• If $n$ is even, then the diagonal crosses the diagonal of the $2$-by-$2$ cells. In the scaled-down version, the diagonal is still a single diagonal starting at $(0, \frac{n}{2})$; otherwise,

• if $n$ is odd, it crosses the corner of the $2$-by-$2$ cells. In the scaled-down version, the diagonal is actually $2$ neighboring diagonals starting at $(0, \frac{n-1}{2})$ and $(0, \frac{n+1}{2})$.

Also, the $2$-by-$2$ cells with values $0$ and $1$ respectively will also have the form:

Illustration

From here we have everything we need to compute the result of the original problem.

Overall, the number of states we have to visit is $\Theta(\log k)$.

Code in Python
• +169

| Write comment?
 » 22 months ago, # |   +2 Quick editorial! Props to arvindf232 for the round now I am very close to orange!
 » 22 months ago, # | ← Rev. 2 →   0 D is missing, any estimated time?UPD: Added, ty
 » 22 months ago, # |   +5 Can this submission for problem F be hacked?
 » 22 months ago, # | ← Rev. 2 →   +20 quick and good editorial!!! As a tester, I think the problemset and solution are very very interesting
 » 22 months ago, # |   +7 Editorial came within 5 mins..... lightning fast editorial. Thanks a lot!
 » 22 months ago, # |   0 Problem D video Editorial
 » 22 months ago, # | ← Rev. 2 →   0 why is my submission getting WA on 3 (problem D)
•  » » 22 months ago, # ^ |   +1 Here's a nice video, by Errichto, that I find helpful in debugging. I take an AC code as a reference, generating small test cases and that does the magic. Hope it helps. :)
•  » » 22 months ago, # ^ |   +1 Take a look at Ticket 16204 from CF Stress for a counter example.
 » 22 months ago, # |   0 why I get wrong answer in D 173227301
•  » » 22 months ago, # ^ |   +1 Take a look at Ticket 16205 from CF Stress for a counter example.
 » 22 months ago, # |   0 C-solution a[i] = (str[i — 1] == '1'); what mdash means here?
•  » » 22 months ago, # ^ |   +3 if (str[i-1] == '1') a[i]=true;else a[i] = false;
 » 22 months ago, # |   0 Thanks for good contest and clear tutorial. I learn a lot in this contest especially problem E and F.
 » 22 months ago, # | ← Rev. 2 →   0 Has anyone tried to submit E using simulated annealing?
 » 22 months ago, # |   0 In problem C, can someone explain why the same solution got TLE at testcase 4 during contest, but wrong answer at testcase 5 after the contest. (side note: It was wrong because of overflow).173206700: TLE at 4 during contest 173239308: Pass at 4, wrong at 5 after contest
•  » » 22 months ago, # ^ |   +1 when you used unordered_map sometime its complexity become O(n^2) for big prime number..that is why your code got tle..
•  » » 22 months ago, # ^ |   +1 https://codeforces.com/blog/entry/62393 you can read this blog...
 » 22 months ago, # | ← Rev. 2 →   0 I liked your proof of the claim in D
 » 22 months ago, # |   +1 I feel like my solution for problem D can be hacked. https://codeforces.com/contest/1734/submission/173205140
•  » » 22 months ago, # ^ |   +2 Its correct because when you are trying to go to right you will take the maximum change which you have got from left side. Similarly for left side.
 » 22 months ago, # |   0 Has anyone tried doing C using BFS? My submission fails 173243305, not able figure out why?
•  » » 22 months ago, # ^ |   0 Consider the following example 1 24 111010101110111010111110 Your code outputs 36 instead of 34 because in BFS 8 is inserted before 6 so cost of 24 is taken as 8 instead of 6.Also do look at editorial's solution. Your approach was very close to the correct apporach
 » 22 months ago, # |   0 why is my 173244335 getting WA on 2 (problem C)
•  » » 22 months ago, # ^ |   0 not looked at your code, but you can do cout the input if test case number is 79 and find out the input and dry run it and find your mistake
•  » » 22 months ago, # ^ |   +1 Take a look at Ticket 16206 from CF Stress for a counter example.
 » 22 months ago, # |   0 Very high quality editorial. Well done.
 » 22 months ago, # |   0 In editorial of E, some extra observation part? How do we get the following conclusion:Let b be the two dimensional difference array of a, that is, bi,j=ai,j−ai−1,j−ai,j−1+ai−1,j−1. Then, the condition becomes sum of any rectangles of b must be nonzero (modn)
•  » » 22 months ago, # ^ |   0 Think of A as a 2D prefix sum array of B.
 » 22 months ago, # |   0 Why this solution does not work for problem C Anyone, please point out my mistake Thank You in Advance #include using namespace std; #define int long long #define nline "\n" const int N = 1e6 + 69, M = 1e9 + 7; char a[N + 1]; int l, r, mid, cand, p, u, v, wt; int n, m, q, t, sm, x, y, k, res; void solve(){ cin >> n; for(int i = 1; i <= n; ++i) cin >> a[i]; bool erased_here[n + 69]; memset(erased_here, 0, sizeof(erased_here)); int res = 0; for(int i = 1; i <= n; ++i){ if(a[i] == '0'){ for(int j = i; j <= n && a[j] == '0'; j += i){ res += i; erased_here[j] = 1; a[j] = '1'; } } else if(erased_here[i]){ for(int j = i + i; j <= n && a[j] == '0'; j += i){ res += i; erased_here[j] = 1; a[j] = '1'; } } } cout << res << nline; } signed main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); int tc = 1; cin>>tc; while(tc--) { solve(); } return 0; } 
•  » » 22 months ago, # ^ |   0 1 9 000010110should be 9 but your code gives 10
 » 22 months ago, # |   0 I have a simple solution for F by search with memorization, but I cannot tell if it's right. Can this submission 173252969 be hacked?
 » 22 months ago, # |   0 Can this 173226289 for problem F be hacked?
 » 22 months ago, # |   0 Can someone please take a look at this D? I reviewed many times, pretty sure it’s O(n), but I get tle. Please tell me what I missed.
•  » » 22 months ago, # ^ |   0 Oh no, not again, can’t delete already. Ещё раз из-за глупости…
 » 22 months ago, # |   0 A pretty optimized $O(N*\sqrt(N))$ solution will pass for problem C.173266505
 » 22 months ago, # |   0 Can anyone explain why this solution 173214401 for problem C gives TLE. I make array of list which contain all possible k for that number initially contain one k which same as its number and i also make an min array which contain current minimum k for that number. so whenever i delete some number from the set i just add that number and its list numbers to the list of next multiple. According to me list contain factors of that number and the factors of 10^6 which is the highest number has 49 factors so acc. to me total no. of operations will be (10^6) * (49) * (10000) = 10^11 which can be done in 2 seconds. Thank you in advance.
•  » » 22 months ago, # ^ |   +1 10^11 takes 100-1000 seconds
•  » » » 22 months ago, # ^ |   0 Okay now I get it in 2 sec 2(10^8) can be done not 10^11 thank you for reply.
•  » » » » 22 months ago, # ^ |   0 yes
 » 22 months ago, # |   +3 In E, why does it matter that n is prime? I don’t see the necessity.
•  » » 22 months ago, # ^ |   0 The condition that "ab = 0 implies a = 0 or b = 0" is true only in prime rings. E.g. if $n=8, (r_1-r_2)(c_1-c_2)=(4)(2)=0$, and $(r_1-r_2) \neq 0, (c_1-c_2) \neq 0$.
•  » » » 22 months ago, # ^ |   0 Got it. Thank you!
 » 22 months ago, # | ← Rev. 2 →   0 for F, "Observe that the i-th character is 1' if and only if i has an odd number of set bits in its binary representation"How to prove this?
•  » » 22 months ago, # ^ |   +1 For each bit with value 1, means the characheter has been flipped once. So if there are odd bits with value 1, the characheter has been flipped odd times, and the origin value is '0', then we can easily know the characheter is '1'.
•  » » » 22 months ago, # ^ |   0 Alternatively, you can find it on the linked wikipedia page
 » 22 months ago, # |   0 Still a long way to go for me. :)
 » 22 months ago, # | ← Rev. 5 →   0 Can anyone please help me with problem D.7 4-1 -2 -4 6 -2 -4 -1I think answer should be YES, But it is given No(in sample test case)my Explaination: let say my currsum=6(as i am on 4th idx){considering 1-based indexing]then i move right my currsum=4(my curr index=5)then i move left my currsum=10(my curr index=4)then i can keep moving right to index 5(currsum=8) then index 6(currsum=2) then index 7(currsum=1), hence i reached destination so answer should be yes?correct me if i am going wrong somewhere.
•  » » 22 months ago, # ^ |   0 Once you consume a slime, you can't consume that again.
 » 22 months ago, # |   0 Can someone please share their logic on how to use segment tree in D? I am just curious. Thank you
 » 22 months ago, # |   +1 Here's my greedy solution for problem D, which is much simpler, but I can't figure out why it works..Mark the area of slimes you have cleared out by keeping track of the left and right boundaries. You begin by expanding toward the left (you take care of expanding toward the right first separately). Either you can reach the end, or you can't. If you can, you win (and print "YES"). If you can't, you settle for going toward the left until you reach the maximum health possible, expanding your territory along the way and then try going toward the right. If you ever fail to expand your territory at all, you stop because it means you'll never reach the end.Do this again for choosing to go right first.Now, each step increases your existing territory. But actually, it may take $O(n)$ time to check whether you are able to reach the end or not since you keep going until you would have fallen to below zero health, which might involve experimentation significantly beyond the edges of the territory. If you only expand the territory by 1 each time but you would be able to reach almost the end if you decided to continue, it might take $O(n)$ for each expansion for an $O(n^2)$ algorithm overall. But either the test cases are too weak or there's something else limiting the time complexity I can't think of. I encourage someone to hack me or figure out why this technique works in subquadratic time.173363257
 » 22 months ago, # |   0 Can anyone help me on 173366132?
 » 22 months ago, # |   0 Could anyone please tell me what's wrong with my solution to problem D? Thanks in advance. https://codeforces.com/contest/1734/submission/173373750
 » 22 months ago, # |   0 I am storing some set of elements in a set . i want to find the previous element of find . eg :- 2 3 4 6 are elements in set can i know the the value of 4 if i had 6 . we can find 6 using s.find() , but how to find previous element of the value we are finding . is it possible ?
•  » » 22 months ago, # ^ |   0 Find iterator (it) of the element you are searching for using find(). Do, it-- for the previous element. Handle the case separately, when the element you are seaching for is at the beginning of the set
•  » » » 22 months ago, # ^ |   0 ok thanks
 » 22 months ago, # | ← Rev. 4 →   0 arvindf232 In problem D, I did the same way as editorial, but mine got TLE on test 9. I have shortened my code as much as possible, kindly please look into that. Code#include #define ll long long using namespace std; #define YES cout << "YES\n" #define Yes cout << "Yes\n" #define NO cout << "NO\n" #define No cout << "No\n" int32_t main() { ios::sync_with_stdio(false); cin.tie(); cout.tie(); int tests; cin >> tests; for (int test = 1; test <= tests; test++) { int n, k; cin >> n >> k; ll p[n]; k--; for(int i=0; i> p[i]; } bool verdict=1; int l=k, r=k; ll chk_l=0, chk_r=0, h=p[k]; while (l>0 and r=0; il--) { cur_h+=(p[il]); if(cur_h<0) { chk_l=h-cur_h; break; } if(il==0) { l=0; break; } if(cur_h>=h) { h=cur_h; l=il; break; } } } if(l==0) { break; } cur_h=h; if(chk_r<=h) { for(int ir=r+1; ir= h) { h = cur_h; r = ir; break; } } } } if(l!=0 and r!=n-1) { verdict=0; } if(verdict) { YES; continue; } NO; } return 0; } 
 » 22 months ago, # |   0 What a shitty and confusing explanation of problem D solution! Can't you explain it in simple words?
 » 21 month(s) ago, # |   0 In problem F, Digit DP solution: "After processing each bit $k$, we should have the following: the number of integers $x$ between $0$ and $⌊\frac{m}{{2^k}}⌋$ inclusive which have the following property:...." I feel difficult to understand why the range is between $0$ and $⌊\frac{m}{{2^k}}⌋$. Anybody explains to me, please. <33
 » 19 months ago, # |   0 I have a much simpler solution for problem ESubmission: 187453380step1: Fill all diagonal elementsstep2: Fill ith row starting from arr[i][i+1] ... with common difference d = i+1That's allexample :B = [1 1 1]` Array Try it out[ 1 2 0 difference = 1 2 1 0 difference = 2 1 1 1 difference = 3 ]
 » 14 months ago, # |   0 /* in this question we have to basically escape our slime so what can we do we can either go to left or to rightnow lets consider if we are going to left then when we should go to left 2 cases are there if we escape or we should increment our healthand that is same for the right if we are going right then either health should be increased or we should escapeso you can maintain two pointers left and right and where you are moving shift the pointers accordingly if left pointer reaches -1 or right pointer reaches n you escaped and change your current health accordingly as you move if you can't move anywhere then the answer is no / / ---------------------------------------- THANK YOU ------------------------------------------->*/
•  » » 14 months ago, # ^ |   0