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
#include<bits/stdc++.h>
using namespace std;
#define MAX_N 1<<17
int n,dat[2*(MAX_N)-1];

//0で初期化
void init(int n_){
  n=1;
  while(n<n_)n*=2;
  for(int i=0;i<2*n-1;i++)dat[i]=0;
}

//k番目の要素の値を返す
int find(int k){
  k+=n-1;
  int ans=dat[k];
  while(k>0){
    k=(k-1)/2;
    ans+=dat[k];
  }
  return ans;
}

//[a,b)の最小値を求める
void update(int a,int b,int x,int k=0,int l=0,int r=n){
  if(r<=a||b<=l)return;
  if(a<=l&&r<=b){
    dat[k]+=x;
  }else{
    update(a,b,x,k*2+1,l,(l+r)/2);
    update(a,b,x,k*2+2,(l+r)/2,r);
  }
}

int main(){
  int q;
  cin>>n>>q;
  init(n);
  for(int i=0;i<q;i++){
    int com,s,t,x;
    cin>>com;
    if(com){
      cin>>t;
      cout<<find(t-1)<<endl;
    }else{
      cin>>s>>t>>x;
      update(s-1,t,x);//[s,t)にxを加算
    }
  }
  return 0;
}

