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:
//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
template<class T> using PQ = priority_queue<T>;
template<class T> using PQG = priority_queue<T, vector<T>, greater<T> >;
const int INF = 100010001;
const ll LINF = (ll)INF*INF*10;
template<typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true);}
template<typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true);}
template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second;}
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << ' ' << p.second;}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, q;
    cin >> n >> q;
    vi a(n);
    int m = sqrt(n);
    vi b((n+m-1)/m);

    auto add = [&](int s, int t, int x) {
        if(s/m == t/m) {
            for(int i = s; i != t; i++) {
                a[i] += x;
            }
        } else {
            for(; s%m; a[s++] += x);
            for(; t%m; a[--t] += x);
            for(s /= m, t /= m; s != t; b[s++] += x);
        }
    };
    auto get = [&](int i) {
        int res = 0;
        res += a[i];
        res += b[i/m];
        return res;
    };

    while(q--) {
        int t;
        cin >> t;
        if(t) {
            int i;
            cin >> i;
            cout << get(i-1) << '\n';
        } else {
            int s, t, x;
            cin >> s >> t >> x;
            add(s-1, t, x);
        }
    }
}