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>
#include <string>
#include <stack>
#include <map>
#include <algorithm>
#include <iomanip>
#include <cmath>
#include <queue>
#include <climits>
using namespace std;


#define MOD 1000000007
using ll = long long;

bool operator<(const pair<int, int>& a, const pair<int, int>& b) {
	if (a.first == b.first) {
		return a.second < b.second;
	}
	return a.first < b.first;
}


const int sqrtN = 512;

struct SqrtDecomposition {
	int N, K;
	vector<int> data;
	vector<int> bucketAdd;
	SqrtDecomposition(int n) : N(n) {
		K = (N + sqrtN - 1) / sqrtN;
		data.assign(K*sqrtN, 0);
		bucketAdd.assign(K, 0);
	}

	// [s, t)
	void add(int s, int t, int x) {
		for (int k = 0; k < K; ++k) {
			int l = k * sqrtN, r = (k + 1) *sqrtN;
			if (r <= s || t <= l) continue;
			if (s <= l && r <= t) {
				bucketAdd[k] += x;
			}
			else {
				for (int i = max(s, l); i < min(t, r); ++i) {
					data[i] += x;
				}
			}
		}
	}

	//[x, y)
	int get(int i) {
		int k = i / sqrtN;
		return data[i] + bucketAdd[k];
	}
};
int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	
	int n, q; cin >> n >> q;
	SqrtDecomposition sd(n);

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