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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<string>
#include<iomanip>
#include<sstream>
#include<map>
#include<cmath>
using namespace std;
#define rep(i,n) for(int i = 0;i < n;i++)
#define req(i,n) for(int i = 1;i <=n;i++)
#define rrep(i,n) for(int i = n -1;i >= 0;i--)
#define ALL(a) a.begin(),a.end()
typedef long long ll;
typedef long double ld;
const ll MOD =  1000000007;
const ll INF =  1LL << 60;
int n, m, x, y,z,w,h, sum= 0, ans;
template<typename A, size_t N, typename T>
void Fill(A(&array)[N], const T & val) {
    std::fill((T*)array, (T*)(array + N), val);
}
template <typename T>
struct BIT {
    int n;             // 要素数
    vector<T> bit[2];  // データの格納先
    BIT(int n_) { init(n_); }
    void init(int n_) {
        n = n_ + 1;
        for (int p = 0; p < 2; p++) bit[p].assign(n, 0);
    }
    void add_sub(int p, int i, T x) {
        for (int idx = i; idx < n; idx += (idx & -idx)) {
            bit[p][idx] += x;
        }
    }
    void add(int l, int r, T x) {  // [l,r) に加算
        add_sub(0, l, -x * (l - 1));
        add_sub(0, r, x * (r - 1));
        add_sub(1, l, x);
        add_sub(1, r, -x);
    }
    T sum_sub(int p, int i) {
        T s(0);
        for (int idx = i; idx > 0; idx -= (idx & -idx)) {
            s += bit[p][idx];
        }
        return s;
    }
    T sum(int i) { return sum_sub(0, i) + sum_sub(1, i) * i; }
};
int main(void) {
    cin >> n >> m; BIT<int> b(n);
    rep(i, m) {
        cin >> x;
        if (x == 0) {
            cin >> y >> z >> w;
            b.add(y, z+1, w);
        }
        else {
            cin >> y;
            cout << b.sum(y) - b.sum(y-1) << endl;
        }
    }
}