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 <iostream>
using namespace std;

template<typename T, int MAX_N>
struct segment_tree_sum {
    T arr1[MAX_N << 1], arr2[MAX_N << 1]; int N;

    segment_tree_sum(int n) {
        N = 1;
        while (N < n) N *= 2;

        for (int i = 0; i < 2*N-1; ++i) {
            arr1[i] = (T)0;
            arr2[i] = (T)0;
        }
    }

    void add(int a, int b, T v) {
        return add(a, b, v, 0, 0, N);
    }

    void add(int a, int b, T v, int k, int l, int r) {
        if (a <= l && r <= b) {
            arr1[k] += v;
        } else if (!(b <= l || r <= a)) {
            arr2[k] += (min(r, b) - max(l, a)) * v;
            const int m = (l+r) / 2;
            add(a, b, v, 2*k+1, l, m);
            add(a, b, v, 2*k+2, m, r);
        }
    }

    T query(int a, int b) {
        return query(a, b, 0, 0, N);
    }

    T query(int a, int b, int k, int l, int r) {
        if (b <= l || r <= a) {
            return (T)0;
        }
        if (a <= l && r <= b) {
            return (r-l) * arr1[k] + arr2[k];
        }
        const int m = (l+r)/2;
        return (min(r, b) - max(l, a)) * arr1[k] + query(a, b, 2*k+1, l, m) + query(a, b, 2*k+2, m, r);
    }
};

int main(void) {
    cin.tie(0); ios::sync_with_stdio(false);
    int N, Q; cin >> N >> Q;
    segment_tree_sum<long long, 201010> segt(N+1);
    for (int j = 0; j < Q; ++j) {
        int c, s, t, x; cin >> c;
        if (c == 0) {
            cin >> s >> t >> x;
            segt.add(s, t+1, x);
        } else {
            cin >> t;
            cout << segt.query(t, t+1) << endl;
        }
    }
    return 0;
}