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>
#define INF 2000000000000000000
#define ll long long
using namespace std;

//=============Segment_Tree_with_Lazy_Propagation====
const int MAX_N = 1 << 17;//要変更
int n;
vector<ll> dat(2 * MAX_N - 1, 0);//要変更
vector<ll> lazy(2 * MAX_N - 1, 0);//要変更

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

void eval(ll k, ll l, ll r) {
  if (lazy.at(k) != 0) {
    dat.at(k) += lazy.at(k);//要変更
    if (r - l > 1) {
      lazy.at(k * 2 + 1) += lazy.at(k);//要変更
      lazy.at(k * 2 + 2) += lazy.at(k);//要変更
    }
    lazy.at(k) = 0;
  }
}

void update(ll a, ll b, ll plus, ll k = 0, ll l = 0, ll r = n) {
  if (r <= a || b <= l) {
    eval(k, l, r);//子供のlazyを更新する必要がある
    return;
  }
  if (a <= l && r <= b) {
    lazy.at(k) += plus;//要変更
    eval(k, l, r);//子供のlazyを更新する必要がある
  }
  else {
    eval(k, l, r);//updateで使う子供のlazeとdat.at(k)の更新
    update(a, b, plus, k * 2 + 1, l, (l + r) / 2);
    update(a, b, plus, k * 2 + 2, (l + r) / 2, r);
    dat.at(k) = dat.at(2 * k + 1) + dat.at(2 * k + 2);//要変更
  }
}

ll query(ll a, ll b, ll k = 0, ll l = 0, ll r = n) {
  if (r <= a || b <= l) {
    return 0;//要変更
  }
  eval(k, l, r);//dat.at(k)と子供のlazy更新のためreturn datとqueryに必要
  if (a <= l && r <= b) {
    return dat.at(k);
  }
  else {
    ll vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
    ll vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
    return vl + vr;//要変更
  }
}
//=================================================

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  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, x);
    }
    else {
      ll index;
      cin >> index;
      cout << query(index - 1, index) << "\n";
    }
  }
}