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>
#define REP(i,n) for (int i = 0; i <(n); ++i)
#define REP2(i,x,n) for (int i = x; i <(n); ++i)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
using namespace std;
using ll = long long;
using P = pair<int,int>;
static const double PI = acos(-1);
static const int INF = 1e9+7;

/*
  RSQ:Range Add Query
  segment tree を利用
  https://www.youtube.com/watch?v=ReGvflPU81c&list=PL3Hpv03CoZ24p5a6qT0LsFKEhiDWxf_B_&index=4
*/

#define SEG_LEN (1 << 18) //10^5 より大きい２進数, index = 1から利用する．
int seg[SEG_LEN * 2];

//one point 取得
long long int get(int ind){
  ind += SEG_LEN;
  long long int sum = 0;
  sum += seg[ind];
  while(true){
    ind /= 2;
    if(ind == 0) break;
    sum += seg[ind];
  }
  return sum;
}

void add(int l, int r, int x){
  l += SEG_LEN;
  r += SEG_LEN;
    while(l < r){
    if(l%2 == 1){
      seg[l] += x;
      l++;
    }
    l /= 2;

    if(r%2 == 1){
      seg[r - 1] += x;
      r--;
    }
    r /= 2;
  }
}

int main(){
  int n, q;
  cin >> n >> q;

  REP(i,q){
    int com;
    cin >> com;
    if(com == 0){//add
      int l, r, x;
      cin >> l >> r >> x;
      add(l, r+1, x);
    }

    if(com == 1){//get
      int ind;
      cin >> ind;
      cout << get(ind) << endl;

    }
  }


  return 0;
}

