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 <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
#define yn {puts("Yes");}else{puts("No");}

// long long version
const int MAX_N = 300005;
ll bit[MAX_N+1];
int nn = MAX_N; //　1-index

void adds(int a, ll w){
    for(int x = a; x <= nn; x += x & -x) bit[x] += w;
}
ll sums(int a){ // bit[1,a]の和を返す.
    ll ret = 0;
    for(int x = a; x > 0; x -= x & -x) ret += bit[x];  
    return ret;
}

int main() {
    int n, q;
    cin >> n >> q;

    rep(loop, q){
        int c;
        cin >> c;
        if(c == 0){
            int s, t, x;
            cin >> s >> t >> x;
            adds(s, x);
            adds(t+1, -x);
        }else{
            int x;
            cin >> x;
            ll res = sums(x);
            cout << res << endl;
        }
    }
    
    return 0;
}
 
 

