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 <bits/stdc++.h>
using namespace std;
using ll = long long;

const ll YMAX = (1ll << 31) - 1;
vector<ll> dat;
int N;

void update(int k, int a) {
  k += N - 1;
  dat[k] += a;
  while (k > 0) {
    k = (k - 1) / 2;
    dat[k] += a;
  }
}

int find(int a, int b, int k, int l, int r) {
  // [a, b) での最小値。[l, r) と節点kが対応
  if (r <= a || b <= l) return 0;
  if (a <= l && r <= b) return dat[k];
  else {
    int vl = find(a, b, k * 2 + 1, l, (l + r) / 2);
    int vr = find(a, b, k * 2 + 2, (l + r) / 2, r);
    return vl + vr;
  }
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  cin >> N;
  ll Q; cin >> Q;
  ll N_ = N;
  N = 1;
  while (N < N_) N *= 2;
  dat = vector<ll>(2 * N - 1, 0);

  for (ll i = 0; i < Q; i++) {
    int q; cin >> q;
    if (q == 0) {
      int s, t, x;
      cin >> s >> t >> x;
      s--;
      update(s, x);
      if (t < N) update(t, -x);
    } else {
      int s; cin >> s;
      int ans = find(0, s, 0, 0, N);
      cout << ans << '\n';
    }
  }

  return 0;
}