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;
template <typename T>
struct range_add_point_value{
	int N;
	vector<T> ST;
	range_add_point_value(int n){
		N = 1;
		while (N < n){
			N *= 2;
		}
		ST = vector<T>(N * 2 - 1, 0);
	}
	range_add_point_value(vector<T> A){
		int n = A.size();
		N = 1;
		while (N < n){
			N *= 2;
		}
		ST = vector<T>(N * 2 - 1, 0);
		for (int i = 0; i < n; i++){
			ST[N - 1 + i] = A[i];
		}
	}
	T operator [](int k){
		k += N - 1;
		T ans = ST[k];
		while (k > 0){
			k = (k - 1) / 2;
			ans += ST[k];
		}
		return ans;
	}
	void add(int L, int R, T x, int i, int l, int r){
		if (R <= l || r <= L){
			return;
		} else if (L <= l && r <= R){
			ST[i] += x;
			return;
		} else {
			int m = (l + r) / 2;
			add(L, R, x, i * 2 + 1, l, m);
			add(L, R, x, i * 2 + 2, m, r);
			return;
		}
	}
	void range_add(int L, int R, T x){
		add(L, R, x, 0, 0, N);
	}
};
int main(){
  int n, q;
  cin >> n >> q;
  range_add_point_value<int> a(n);
  for (int i = 0; i < q; i++){
    int T;
    cin >> T;
    if (T == 0){
      int s, t, x;
      cin >> s >> t >> x;
      a.range_add(s - 1, t, x);
    }
    if (T == 1){
      int t;
      cin >> t;
      cout << a[t - 1] << endl;
    }
  }
}