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;

int n, m, a[100005], c[100005];

int lowbit(int x)
{
	return x & (-x);
}

void updata(int i, int k)
{
	while (i <= n){
		c[i] += k;
		i += lowbit(i);
	}
}

int getsum(int i)
{
	int res = 0;
	while(i > 0){
		res += c[i];
		i -= lowbit(i);
	}
	return res;
}

int main()
{
	cin >> n >> m;
	for (int i = 1; i <= n; ++i)
	{
		updata(i, a[i] - a[i - 1]);
	}
	while (m--){
		int ch, x, y, k;
		scanf("%d%d", &ch, &x);
		if (!ch){
			scanf("%d%d", &y, &k);
			updata(x, k);
			updata(y + 1, -k);
		}else{
			printf("%d\n", getsum(x));
		}
	}
	return 0;
}