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;
using ll=long long;
using Graph=vector<vector<int>>;
using pp=pair<int,pair<int,int>>;
#define MAX 200003
#define MOD 1000000007
#define INF 1000000000000000000

int main(){
  int n,q;
  cin>>n>>q;
  int k=sqrt(n);
  int l=(n+k-1)/k;

  vector<ll> a(n,0);
  vector<ll> b(l,0);

  vector<ll> ans;
  for(int i=0;i<q;i++){
    int c;
    cin>>c;
    if(c==0){
      int s,t;
      ll x;
      cin>>s>>t>>x;
      s--;
      while(s<t&&s%k>0){
        a.at(s)+=x;
        s++;
      }
      while(s<t&&t%k>0){
        t--;
        a.at(t)+=x;
      }
      while(s<t){
        b.at(s/k)+=x;
        s+=k;
      }
    }else{
      int x;
      cin>>x;
      x--;
      ans.push_back(a.at(x)+b.at(x/k));
    }
  }

  for(int i=0;i<ans.size();i++){
    cout<<ans.at(i)<<endl;
  }
}

