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 N 100010
using namespace std;
struct no{
    int v;//value
    int l;//left
    int r;//right
}t[N*4];//存储的是变化值
int n, m;
void build(int i, int l, int r){
    t[i] = {0, l, r};
    if(l == r)return;;
    int mid = (l+r)/2;
    build(2*i, l, mid);
    build(2*i+1, mid+1, r);
}
void pls(int i, int l, int r, int k){
    if(l <= t[i].l && t[i].r <= r){
        t[i].v += k;
        return ;
    }
    if(l <= t[2*i].r){
        pls(2*i, l ,r, k);
    }
    if(t[2*i+1].l <= r){
        pls(2*i+1, l, r, k);
    }
}
int ans = 0;
void query(int i, int p){
    ans += t[i].v;
    if(t[i].l == t[i].r) return ;
    if(p <= t[2*i].r){
        query(2*i, p);
    }
    if(t[2*i+1].l <= p){
        query(2*i+1, p);
    }
}
int main() {
    scanf("%d %d", &n, &m);
    build(1, 1, n);
    for(int i = 0; i < m; i++){
        int op;
        scanf("%d", &op);
        if(op == 0){//区间加
            int b, c, k;
            scanf("%d %d %d", &b, &c, &k);
            pls(1, b, c, k);
        }else if(op == 1){//单点查
            int p;
            scanf("%d", &p);
            ans = 0;
            query(1, p);
            printf("%d\n", ans);
        }
    }
    return 0;
}
