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 rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<typename T>
struct BIT{
    private:
        vector<T> array;
        const int n;
    public:
        BIT(int _n) : array(_n + 1, 0), n(_n) {}
        T sum(int i){
            T s = 0;
            while(i > 0){
                s += array[i];
                i -= i & -i;
            }
            return s;
        }
        T sum(int i, int j){
            T ret_i = sum(i-1);
            T ret_j = sum(j);
            return ret_j - ret_i;
        }
        void add(int i, T x){
            while(i <= n){
                array[i] += x;
                i += i & -i;
            }
        }
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, q;
    cin >> n >> q;
    BIT<int> bit0(n), bit1(n);
	int k, s, t, x;
    rep(i,q){
		cin >> k;
        if(k == 0){
            cin >> s >> t >> x;
            bit0.add(s, -x * (s - 1));
            bit1.add(s, x);
            bit0.add(t+1, x * t);
            bit1.add(t+1, -x);
        }
        else{
			cin >> s;
			t = s;
            int ans = 0;
            ans += bit0.sum(t) + bit1.sum(t) * t;
            ans -= bit0.sum(s - 1) + bit1.sum(s - 1) * (s - 1);
            cout << ans << endl;
        }
    }
}