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;

typedef long long ll;
static const int MAX_N = 1 << 17;

struct StarrySky {
	int datm[2 * MAX_N], data[2 * MAX_N], n;
	StarrySky() {}
	StarrySky(int n_) { init(n_); }

	void init(int n_) {
		n = 1;
		while (n<n_) n *= 2;
		for (int i = 0; i<n * 2; i++) datm[i] = data[i] = 0;
	}

	void add(int a, int b, int x, int k, int l, int r) {
		if (r <= a || b <= l) return;
		if (a <= l && r <= b) {
			data[k] += x;
			return;
		}
		add(a, b, x, k * 2 + 1, l, (l + r) / 2);
		add(a, b, x, k * 2 + 2, (l + r) / 2, r);
		datm[k] = max(datm[k * 2 + 1] + data[k * 2 + 1], datm[k * 2 + 2] + data[k * 2 + 2]);
	}

	int get(int a, int b, int k, int l, int r) {
		if (r <= a || b <= l) return 0;
		if (a <= l && r <= b) return datm[k] + data[k];
		int vl = get(a, b, k * 2 + 1, l, (l + r) / 2);
		int vr = get(a, b, k * 2 + 2, (l + r) / 2, r);
		return max(vl, vr) + data[k];
	}

	void add(int a, int b, int x) {
		add(a, b, x, 0, 0, n);
	}

	int get(int a, int b) {
		return get(a, b, 0, 0, n);
	}

};

int main() {
	int n, q;
	cin >> n >> q;

	StarrySky ss(n);

	for (int i = 0; i < q; i++) {
		int c, s, t, x;
		cin >> c;
		if (c) {
			cin >> x;
			cout << ss.get(x - 1, x) << endl;
		}
		else {
			cin >> s >> t >> x;
			ss.add(s - 1, t, x);
		}
	}
	return 0;
}