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 INF 5000000000000000000
#define ll long long
#define pll pair<ll, ll>
using namespace std;
//=============segment_tree============================
const int MAX_N = 1 << 17;
int n;
vector<ll> dat(2 * MAX_N - 1, 0);

void init(ll n_) {
  n = 1;
  while (n < n_) {
    n *= 2;
  }
}

void update(ll a, ll b, ll k, ll l, ll r, ll s) {
  if (r <= a || b <= l) {
    return;
  }
  if (a <= l && r <= b) {
    dat.at(k) += s;
  }
  else {
    update(a, b, k * 2 + 1, l, (l + r) / 2, s);
    update(a, b, k * 2 + 2, (l + r) / 2, r, s);
  }
}

ll query(ll k) {
  k += n - 1;
  ll res = dat.at(k);
  while (k > 0) {
    k = (k - 1) / 2;
    res += dat.at(k);
  }
  return res;
}
//=================================================

int main() {
  ll N, Q;
  cin >> N >> Q;
  init(N);
  for (ll i = 0; i < Q; ++i) {
    ll q;
    cin >> q;
    if (q == 0) {
      ll s, t, x;
      cin >> s >> t >> x;
      update(s - 1, t, 0, 0, n, x);
    }
    else {
      ll index;
      cin >> index;
      cout << query(index - 1) << "\n";
    }
  }
}

