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;


// 区間加算　一点取得 RAQ 平方分割で実装
template <typename T>
struct RAQ{// 0-index
	int n,rn,bn; // rn個のかたまりをbn個
	vector<T> data,bucket;
	RAQ(int n_){init(n_);}
	void init(int n_){
		n = n_;
		rn = sqrt(n);
		bn = ceil((double)n/rn);
		data.resize(n,0);
		bucket.resize(bn,0);
	}
	T get(int i){
		return data[i]+bucket[i/rn];
	}
	void add(int s,int t,T x){ // s~tにxを加算
		for(int i=s;i<=t;++i){
			if(i%rn==0 && (i+rn-1)<=t){
				bucket[i/rn] += x;
				i += rn-1;
			}else{
				data[i] += x;
			}
		}
	}
};


int main(){
	
	int n,q;
	cin >> n >> q;
	RAQ<ll> raq(n);
	while(q--){
		int com,s,t;
		ll x;
		cin >> com;
		if(com){
			cin >> s;
			cout << raq.get(s-1) << "\n";
		}else{
			cin >> s >> t >> x;
			raq.add(s-1,t-1,x);
		}
	}
}