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<iomanip>
#include <cmath>
#include <climits>
#include <algorithm>
#include <stdio.h>
#include <vector>
#include <queue>
#include <tuple>
#include <map>
#include <list>
#include <string>
#include <numeric>
#include <utility>
#include <cfloat>
#include <set>

using namespace std;

int sqrtN = 512;
struct SqrtDecomposition{
    int N, K;
    vector <long long> data;
    vector <long long> bucketAdded;
    
    SqrtDecomposition(int n){
        N = n;
        K = (N + sqrtN - 1) / sqrtN;
        data.assign(N, 0);
        bucketAdded.assign(K, 0);
    }

    long long get(int x){
        return data[x] + bucketAdded[x / sqrtN];
    }

    void add(int x, int y, int a){
        if(y - x < sqrtN){
            for(int i = x; i < y; i++){
                data[i] += a;
            }
            return;
        }
        for(int i = x / sqrtN + 1; i < y / sqrtN; i++){
            bucketAdded[i] += a;
        }

        for(int i = x; i < (x / sqrtN + 1) * sqrtN; i++){
            data[i] += a;
        }
        for(int i = y / sqrtN * sqrtN; i < y; i++){
            data[i] += a;
        }
    }
    
};

int main(){
    int n;
    int q;
    cin >> n >> q;
    SqrtDecomposition sq(n);
    for(int i = 0; i < q; i++){
        int com;
        int x;
        int y;
        int a;
        cin >> com;
        if(com == 0){
            cin >> x >> y >> a;
            sq.add(x - 1, y, a);
        } else {
            cin >> x;
            cout << sq.get(x - 1) << endl;
        }
    }
    return 0;
}