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VK Cup 2012 Round 3 |
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Finished |

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E. Polycarpus and Tasks

time limit per test

3 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputPolycarpus has many tasks. Each task is characterized by three integers *l*_{i}, *r*_{i} and *t*_{i}. Three integers (*l*_{i}, *r*_{i}, *t*_{i}) mean that to perform task *i*, one needs to choose an integer *s*_{i} (*l*_{i} ≤ *s*_{i}; *s*_{i} + *t*_{i} - 1 ≤ *r*_{i}), then the task will be carried out continuously for *t*_{i} units of time, starting at time *s*_{i} and up to time *s*_{i} + *t*_{i} - 1, inclusive. In other words, a task is performed for a continuous period of time lasting *t*_{i}, should be started no earlier than *l*_{i}, and completed no later than *r*_{i}.

Polycarpus's tasks have a surprising property: for any task *j*, *k* (with *j* < *k*) *l*_{j} < *l*_{k} and *r*_{j} < *r*_{k}.

Let's suppose there is an ordered set of tasks *A*, containing |*A*| tasks. We'll assume that *a*_{j} = (*l*_{j}, *r*_{j}, *t*_{j}) (1 ≤ *j* ≤ |*A*|). Also, we'll assume that the tasks are ordered by increasing *l*_{j} with the increase in number.

Let's consider the following recursive function *f*, whose argument is an ordered set of tasks *A*, and the result is an integer. The function *f*(*A*) is defined by the greedy algorithm, which is described below in a pseudo-language of programming.

- Step 1. ,
*ans*= 0. - Step 2. We consider all tasks in the order of increasing of their numbers in the set
*A*. Lets define the current task counter*i*= 0. - Step 3. Consider the next task:
*i*=*i*+ 1. If*i*> |*A*| fulfilled, then go to the 8 step. - Step 4. If you can get the task done starting at time
*s*_{i}= max(*ans*+ 1,*l*_{i}), then do the task*i*:*s*_{i}= max(*ans*+ 1,*l*_{i}),*ans*=*s*_{i}+*t*_{i}- 1, . Go to the next task (step 3). - Step 5. Otherwise, find such task , that first, task
*a*_{i}can be done at time*s*_{i}= max, and secondly, the value of is positive and takes the maximum value among all*b*_{k}that satisfy the first condition. If you can choose multiple tasks as*b*_{k}, choose the one with the maximum number in set*A*. - Step 6. If you managed to choose task
*b*_{k}, then , . Go to the next task (step 3). - Step 7. If you didn't manage to choose task
*b*_{k}, then skip task*i*. Go to the next task (step 3). - Step 8. Return
*ans*as a result of executing*f*(*A*).

Polycarpus got entangled in all these formulas and definitions, so he asked you to simulate the execution of the function *f*, calculate the value of *f*(*A*).

Input

The first line of the input contains a single integer *n* (1 ≤ *n* ≤ 10^{5}) — the number of tasks in set *A*.

Then *n* lines describe the tasks. The *i*-th line contains three space-separated integers *l*_{i}, *r*_{i}, *t*_{i} (1 ≤ *l*_{i} ≤ *r*_{i} ≤ 10^{9}, 1 ≤ *t*_{i} ≤ *r*_{i} - *l*_{i} + 1) — the description of the *i*-th task.

It is guaranteed that for any tasks *j*, *k* (considering that *j* < *k*) the following is true: *l*_{j} < *l*_{k} and *r*_{j} < *r*_{k}.

Output

For each task *i* print a single integer — the result of processing task *i* on the *i*-th iteration of the cycle (step 3) in function *f*(*A*). In the *i*-th line print:

- 0 — if you managed to add task
*i*on step 4. - -1 — if you didn't manage to add or replace task
*i*(step 7). -
*res*_{i}(1 ≤*res*_{i}≤*n*) — if you managed to replace the task (step 6):*res*_{i}equals the task number (in set*A*), that should be chosen as*b*_{k}and replaced by task*a*_{i}.

Examples

Input

5

1 8 5

2 9 3

3 10 3

8 11 4

11 12 2

Output

0 0 1 0 -1

Input

13

1 8 5

2 9 4

3 10 1

4 11 3

8 12 5

9 13 5

10 14 5

11 15 1

12 16 1

13 17 1

14 18 3

15 19 3

16 20 2

Output

0 0 0 2 -1 -1 0 0 0 0 7 0 12

Codeforces (c) Copyright 2010-2024 Mike Mirzayanov

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