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>
#include <vector>

using namespace std;

const int DEFAULT = 0;

//Regular lazy propagation
struct SegmentTree {
    int n;
    vector<int> heap;
    
    SegmentTree(int n): n(n) {
        heap.assign(4 * n, DEFAULT);
    }
    
    void update(int nodeId, int l, int r, int ql, int qr, int v) {
        if (qr < l || ql > r) {
            return;
        }
        else if (ql <= l && r <= qr) {
            heap[nodeId] += v;
        }
        else {
            pushdown(nodeId);
            int m = l + (r - l)/2;
            update(nodeId * 2, l, m, ql, qr, v);
            update(nodeId * 2 + 1, m + 1, r, ql, qr, v);
        }
    }
    
      void pushdown(int nodeId) {
        if (heap[nodeId] > 0) {
            heap[nodeId * 2] += heap[nodeId];
            heap[nodeId * 2 + 1] += heap[nodeId];
            heap[nodeId] = 0;
        }
    }
    
    int query(int nodeId, int l, int r, int q) {
        if (q < l || q > r) {
            return 0;
        }
        if (l == r) {
            return heap[nodeId];
        } 
        else {
            pushdown(nodeId);
            int m = l + (r - l)/2;
            return query(nodeId * 2, l, m, q) + query(nodeId * 2 + 1, m + 1, r, q);
        }
    }
    
};

int main() {
    int n, q;
    cin >> n >> q;
    SegmentTree st(n);
    for (int i = 0; i < q; i++) {
        int cmd;
        cin >> cmd;
        if (cmd == 0) {
            int ql, qr, v;
            cin >> ql >> qr >> v;
            st.update(1, 1, n, ql, qr, v);
//            for (int i = 0; i < 4 * n; i++) {
//                cout << i << ":" << st.heap[i] << endl; 
//            }
        }
        else {
            int pos;
            cin >> pos;
            cout << st.query(1, 1, n, pos) << endl;
        }
    }
    
    return 0;
}