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
#include <iostream>
#include <list>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <algorithm>
#include <cmath>
#include <string>
using namespace std;
typedef long long lli;
typedef vector<lli> vll;
typedef vector<bool> vbl;
typedef vector<vector<lli> > mat;
typedef vector<vector<bool> > matb;
typedef vector<string> vst;
typedef pair<lli,lli> pll;
typedef pair<double,double> pdd;

template<class T>
class bit : public vector<T> {
private:
    lli n;
public:
    bit():vector<T>(){}
    bit(lli n):vector<T>(n+1,0){
        this->n = n;
    }
    T sum(lli i){
        T s = 0;
        for(;i > 0;i -= i & -i) s += this->at(i);
        return s;
    }
    void add(lli i,T x){
        for(;i <= n;i += i & -i) this->at(i) += x;
    }
};

bit<lli> b;

lli n,q;
lli c,s,t,x;

int main(){
    cin >> n >> q;
    b = bit<lli> (n+1);
    for(lli i = 0;i < q;i++){
        cin >> c;
        if(c == 0){
            cin >> s >> t >> x;
            b.add(s,x);
            b.add(t+1,-x);
        }else if(c == 1){
            cin >> x;
            cout << b.sum(x) << endl;
        }
    }
    return 0;
}