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 <iostream>
using namespace std;

const int DEPTH = 17;
struct RAQ {
	int d[1 << DEPTH + 1];
	
	RAQ() {
		for (int i = 0; i < 1 << DEPTH + 1; i++) {
			d[i] = 0;
		}
	}
	
	void add(int l, int r, int x, int a = 0, int b = (1 << DEPTH), int id = 0) {
		if (a >= r || b <= l) return;
		if (l <= a && b <= r) {
			d[id] += x;
			return;
		}
		add(l, r, x, a, (a + b) / 2, id * 2 + 1);
		add(l, r, x, (a + b) / 2, b, id * 2 + 2);
	}
	
	int query(int pos) {
		pos += (1 << DEPTH) - 1;
		
		int ret = d[pos];
		while (pos > 0) {
			pos = (pos - 1) / 2;
			ret += d[pos];
		}
		return ret;
	}
};

RAQ raq;

int main() {
	int n, q;
	cin >> n >> q;
	
	for (int i = 0; i < q; i++) {
		int type;
		cin >> type;
		
		if (type == 0) {
			int s, t, x;
			cin >> s >> t >> x;
			raq.add(s, t + 1, x);
		}
		else {
			int pos;
			cin >> pos;
			cout << raq.query(pos) << endl;
		}
	}
	
	return 0;
}
