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;
using ll = long long;

struct LazySegmentTree {
    int size;
    vector<ll> node, lazy;

    LazySegmentTree(int n) {
        size = 1;
        while (size < n) size <<= 1;
        node.resize(2 * size);
        lazy.resize(2 * size);
    }

    void eval(int k) {
        if (lazy[k] == 0) return;
        node[k] += lazy[k];
        if (k < size) {
            lazy[2 * k] += lazy[k] / 2;
            lazy[2 * k + 1] += lazy[k] / 2;
        }
        lazy[k] = 0;
    }

    void update(int a, int b, ll x, int k = 1, int l = 0, int r = -1) {
        if (r == -1) r = size;
        eval(k);
        if (r <= a || b <= l) return;
        if (a <= l && r <= b) {
            lazy[k] = (r - l) * x;
            eval(k);
        } else {
            int m = (l + r) / 2;
            update(a, b, x, 2 * k, l, m);
            update(a, b, x, 2 * k + 1, m, r);
            node[k] = node[2 * k] + node[2 * k + 1];
        }
    }

    ll query(int i, int k = 1, int l = 0, int r = -1) {
        if (r == -1) r = size;

        eval(k);
        int m = (l + r) / 2;
        if (k >= size && l == i) return node[k];
        if (l <= i && i < m) return query(i, 2 * k, l, m);
        else return query(i, 2 * k + 1, m, r);
    }
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int n, q;
    cin >> n >> q;
    LazySegmentTree st(n);
    for (int j = 0; j < q; j++) {
        int type;
        cin >> type;
        if (type == 0) {
            int s, t, x;
            cin >> s >> t >> x;
            st.update(s - 1, t, x);
        } else {
            int i;
            cin >> i;
            cout << st.query(i - 1) << '\n';
        }
    }
}