Codeforces Round 223 (Div. 1) |
---|

Finished |

Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

graphs

implementation

*2400

No tag edit access

The problem statement has recently been changed. View the changes.

×
B. Sereja and Tree

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputSereja adores trees. Today he came up with a revolutionary new type of binary root trees.

His new tree consists of *n* levels, each vertex is indexed by two integers: the number of the level and the number of the vertex on the current level. The tree root is at level 1, its index is (1, 1). Here is a pseudo code of tree construction.

//the global data are integer arrays cnt[], left[][], right[][]

cnt[1] = 1;

fill arrays left[][], right[][] with values -1;

for(level = 1; level < n; level = level + 1){

cnt[level + 1] = 0;

for(position = 1; position <= cnt[level]; position = position + 1){

if(the value of position is a power of two){ // that is, 1, 2, 4, 8...

left[level][position] = cnt[level + 1] + 1;

right[level][position] = cnt[level + 1] + 2;

cnt[level + 1] = cnt[level + 1] + 2;

}else{

right[level][position] = cnt[level + 1] + 1;

cnt[level + 1] = cnt[level + 1] + 1;

}

}

}

After the pseudo code is run, cell cnt[level] contains the number of vertices on level *level*. Cell left[level][position] contains the number of the vertex on the level *level* + 1, which is the left child of the vertex with index (*level*, *position*), or it contains -1, if the vertex doesn't have a left child. Similarly, cell right[level][position] is responsible for the right child. You can see how the tree with *n* = 4 looks like in the notes.

Serja loves to make things complicated, so he first made a tree and then added an empty set *A*(*level*, *position*) for each vertex. Then Sereja executes *m* operations. Each operation is of one of the two following types:

- The format of the operation is "1
*t**l**r**x*". For all vertices*level*,*position*(*level*=*t*;*l*≤*position*≤*r*) add value*x*to set*A*(*level*,*position*). - The format of the operation is "2
*t**v*". For vertex*level*,*position*(*level*=*t*,*position*=*v*), find the union of all sets of vertices that are in the subtree of vertex (*level*,*position*). Print the size of the union of these sets.

Help Sereja execute the operations. In this problem a set contains only distinct values like std::set in C++.

Input

The first line contains integers *n* and *m* (1 ≤ *n*, *m* ≤ 7000).

Next *m* lines contain the descriptions of the operations. The operation of the first type is given by five integers: 1 *t* *l* *r* *x* (1 ≤ *t* ≤ *n*; 1 ≤ *l* ≤ *r* ≤ *cnt*[*t*]; 1 ≤ *x* ≤ 10^{6}). The operation of the second type is given by three integers: 2 *t* *v* (1 ≤ *t* ≤ *n*; 1 ≤ *v* ≤ *cnt*[*t*]).

Output

For each operation of the second type, print the answer on a single line.

Examples

Input

4 5

1 4 4 7 1

1 3 1 2 2

2 1 1

2 4 1

2 3 3

Output

2

0

1

Note

You can find the definitions that are used while working with root trees by this link: http://en.wikipedia.org/wiki/Tree_(graph_theory)

You can see an example of a constructed tree at *n* = 4 below.

Codeforces (c) Copyright 2010-2023 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Sep/23/2023 21:04:39 (i2).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|