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;
typedef long long ll;

struct RAQ {
    int sz;
    vector<int> dat;
    inline RAQ(int n) {
        sz = 1;
        while(sz < n)
            sz <<= 1;
        dat.assign(sz << 1, 0);
    }
    inline void add(int l, int r, int x) {
        for(l += sz, r += sz; l < r; l >>= 1, r >>= 1) {
            if(l & 1) dat[l++] += x;
            if(r & 1) dat[--r] += x;
        }
    }
    inline int get(int k) {
        int rtn = 0;
        for(k += sz; k; k >>= 1)
            rtn += dat[k];
        return rtn;
    }
};

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int N, Q;
    cin >> N >> Q;

    RAQ seg(N);

    int c, x, y, z;
    while(Q--) {
        cin >> c;
        if(c) {
            cin >> x;
            cout << seg.get(x - 1) << "\n";
        } else {
            cin >> x >> y >> z;
            seg.add(x - 1, y, z);
        }
    }
}