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<fstream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<stack>
#include<climits>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#include<complex>
#include<typeinfo>
using namespace std;
typedef unsigned long long ull;
const int N_MAX = 200000;

struct raqSB{
    int dataSize;
    int bucketSize;
    ull data[N_MAX];
    ull bucket[N_MAX];

    raqSB(const int n){
        bucketSize = (int)sqrt(n) + 1;
        dataSize = bucketSize * bucketSize;
        for (int i=0;i<N_MAX;i++){
            data[i] = 0ull;
            bucket[i] = 0ull;
        }
    }

    void add(int s, int t, int x){
        int ss = s/bucketSize;
        int tt = t/bucketSize;
        if(ss==tt){
            for (int i=s;i<t+1;i++){
                data[i] += x;
            }
        }else{
            for(int i=s;i<(ss+1)*bucketSize;i++){
                data[i] += x;
            }
            for(int i=tt*bucketSize;i<t+1;i++){
                data[i] += x;
            }
            for(int i=ss+1;i<tt;i++){
                bucket[i] += x;
            }
        }
    }

    ull get(int i){
        return data[i] + bucket[i/bucketSize];
    }
};

int main(){
    int n, q, s, t, i, x;
    cin >> n >> q;
    bool qType;
    raqSB raqsb(n);
    for (int j=0;j<q;j++){
        cin >> qType;
        if(qType){
            cin >> i;
            cout << raqsb.get(i) << endl;
        }else{
            cin >> s >> t >> x;
            raqsb.add(s, t, x);
        }
    }
    return 0;
}