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 <stdio.h>
typedef long long ll;

int main(void) {
  ll i, j, k, n, q, c, s, t, x;
  scanf("%lld%lld", &n, &q);
  ll f = 1, seg[4 * n];
  while(f < n) f *= 2;
  for(i = f; i < f * 2; ++i) seg[i] = 0;
  for(i = 0; i < f; ++i) seg[i] = 0;
  for(i = 1; i <= q; ++i) {
    scanf("%lld", &c);
    if(!c) {
      scanf("%lld%lld%lld", &s, &t, &x);
      s--, t--;
      ll r = s + f;
      while(1) {
        ll las, ra = 1;
        while(!(r % 2)) r /= 2;
        while(ra <= r) ra *= 2;
        ra /= 2;
        las = f / ra * (r + 1 - ra) - 1;
        while(las > t) {
          r *= 2;
          ra *= 2;
          las = f / ra * (r + 1 - ra) - 1;
        }
        seg[r] += x;
        //printf("r%lld ra%lld las%lld\n", t, ra, las);
        if(las == t) break;
        r++;
      }
    } else {
      scanf("%d", &s);
      s--;
      int now = s + f, ans = 0;
      while(now) {
         ans += seg[now];
         now /= 2;
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}
