Problem: Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:

* add(s, t, x): add x to as, as+1, ..., at.
* get(i): output the value of ai.



Note that the initial values of ai (i = 1, 2, . . . , n) are 0.

Constraints

* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 1 ≤ s ≤ t ≤ n
* 1 ≤ i ≤ n
* 0 ≤ x ≤ 1000

Input


n q
query1
query2
:
queryq


In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:


0 s t x


or


1 t


The first digit represents the type of the query. '0' denotes add(s, t, x) and '1' denotes get(i).

Output

For each get operation, print the value.

Examples

Input

3 5
0 1 2 1
0 2 3 2
0 3 3 3
1 2
1 3


Output

3
5


Input

4 3
1 2
0 1 4 1
1 2


Output

0
1

Solution (in CPP):

Solution:
//区間更新、一点取得を出来るBIT
#include<bits/stdc++.h>
using namespace std;
struct BinaryIndexedTree{
  int N;
  vector<int> bit;  
  //全て0で初期化
  BinaryIndexedTree(int size){
    bit.resize(size+1,0);
    N=size+1;
  }

  //区間[a,b]の所v[a],v[a+1], v[b]に値wを加える。O(log n)
  void add(int a,int b,int w){
    //bit[a]にwを、bit[b+1]に-wを加える
    for(int x=a;x<=N;x+=(x & -x)) bit[x]+=w;
    for(int x=b+1;x<=N;x+=(x & -x)) bit[x]-=w;
    return;
  }

  //v[a]の値を求める。
  int sum(int a){
    //和 bit[1]+bit[2]+  +bit[a]を求める。
    int ret=0;
    for(int x=a;x>0;x-=(x & -x)) ret+=bit[x];
    return ret;
  }
};
//AOJ DSL 2 E Range Add Query(RAQ) で動作確認しました。
int main(){
    int n,q;
    cin>>n>>q;
    BinaryIndexedTree BIT(n+5);
    int que,s,t,x,a;
    for(int i=0;i<q;i++){
      cin>>que;
      if(que==0){
        cin>>s>>t>>x;
        BIT.add(s,t,x);
      }
      else{
        cin>>a;
        cout<<BIT.sum(a)<<'\n';
      }
    }
    return(0);
}