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 <stdio.h>
#include <fstream>
#include <algorithm>
#include <string>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <limits.h>
#include <math.h>

#define rep(i,n) for (int i = 0; (i) < (n); ++ (i))

using namespace std;

const long long sqrtN = 512;
struct SqrtDecomposition {
  long long N, K;
  vector<long long> data;
  vector<long long> bucketAdd;
  SqrtDecomposition(long long n) : N(n) {
    K = (N + sqrtN - 1) / sqrtN;
    data.assign(K * sqrtN, 0);
    bucketAdd.assign(K, 0);
  }
  void add(long long s, long long t, long long x) {
    for (long long k = 0; k < K; ++k) {
      long long l = k * sqrtN, r = (k + 1) * sqrtN;
      if (r <= s || t <= l)
        continue;
      if (s <= l && r <= t) {
        bucketAdd[k] += x;
      } else {
        for (long long i = max(s, l); i < min(t, r); ++i) {
          data[i] += x;
        }
      }
    }
  }
  long long get(long long i) {
    long long k = i / sqrtN;
    return data[i] + bucketAdd[k];
  }
};

int main() {
    long long n,q,que,s,t,x;
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> n >> q;
    SqrtDecomposition qq(n);
    rep(i,q) {
        cin >> que;
        if(que) {
            cin >> x;
            cout << qq.get(x) << endl;
        } else {
            cin >> s >> t >> x;
            qq.add(s,t+1,x);
        }
    }
    return 0;
}