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"
#include <array>
using namespace std;

struct RAQ{
private:
	struct Node{
		long long sum; long long buf;
		Node(long long s, long long b) : sum(s), buf(b){}
	};
	vector<Node> node;
	long long defval;
	int N;
	int n;
public:
	void init(int n, long long defval = 0){
		N = 1;
		while(N < n)N<<=1;
		this->n = n;
		this->defval = defval;
		node.assign(N*2-1, Node(defval, 0));
	}
	long long query(int a, int b, int l = 0, int r = -1, int i = 0){
		if(r == -1) r = N;
		if(b <= l || r <= a) return defval;
		if(a <= l && r <= b) return node[i].sum;
		int rag = min(r, b) - max(l, a);
		return query(a,b,l,(l+r)/2,i*2+1) + query(a,b,(l+r)/2,r,i*2+2) + node[i].buf * rag;
	}
	void update(int a, int b, long long v, int l = 0, int r = -1, int i = 0){
		if(r == -1) r = N;
		if(b <= l || r <= a) return;
		if(a <= l && r <= b) { node[i].buf += v; node[i].sum += v * (r-l); return; }
		
		update(a,b,v,l,(l+r)/2,i*2+1);
		update(a,b,v,(l+r)/2,r,i*2+2);

		node[i].sum = node[i*2+1].sum + node[i*2+2].sum + node[i].buf * (r-l);
	}

};


int main() {
	int n, q; cin >> n >> q;
	RAQ tr; tr.init(n, 0);

	vector<long long> ans;

	for(int i = 0; i < q; i++){
		int c; cin >> c;
		if(c == 0){
			int s,t,x; cin >> s >> t >> x;
			tr.update(s-1, t, x);
		}
		else {
			int i; cin >> i;
			ans.push_back(tr.query(i-1, i));
		}
	}

	for(int i = 0; i < ans.size(); i++){
		cout << ans[i] << endl;
	}

    return 0;
}