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:
#include<bits/stdc++.h>
using namespace std;
#define MAX 100005
  
// 262144 = 2^18
int tree[262144]; 
  
int dfs(int a,int b,int k,int l,int r){
  if(r<a || b<l)return 0;
  if(a<=l && r<=b)return tree[k];
  int m=(l+r)/2;
  return dfs(a,b,k*2+1,l,m)+dfs(a,b,k*2+2,m+1,r);
}
  
int getSum(int a,int b){
  // 131072 = 2^17
  return dfs(a,b,0,0,131071);
}
  
void update(int i,int x){
  // 131072 = 2^17
  i+=(131072-1);
  tree[i]+=x;
  while(i){
    i=(i-1)/2;
    tree[i]+=x;
  }
}

int n,q;
  
int main(){
  scanf("%d %d",&n,&q);
  while(q--){
    int type;
    scanf("%d",&type);
    if(type==0){
      int l,r,x;
      scanf("%d %d %d",&l,&r,&x);
      update(l,x);
      update(r+1,-x);
      
    }else{
      int target;
      scanf("%d",&target);
      int ans=getSum(0,target);
      printf("%d\n", ans);
    }
  }
  return 0;
}