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>
using namespace std;
#define ll long long
int n, q, n2;
vector<ll> node, lazy;

void eval(int k, int l, int r){
    node[k] += lazy[k];
    if(r - l > 1){
        lazy[k * 2 + 1] += lazy[k] / 2;
        lazy[k * 2 + 2] += lazy[k] / 2;
    }
    lazy[k] = 0;
}

void add(int a, int b, ll x, int k = 0, int l = 0, int r = -1){
    if(r < 0) r = n2;

    eval(k, l, r);

    if(b <= l || r <= a) return;
    if(a <= l && r <= b){
        lazy[k] += x *(ll)(r - l);
        eval(k, l, r);
        return;
    }
    add(a, b, x, k * 2 + 1, l, (l + r) / 2);
    add(a, b, x, k * 2 + 2, (l + r) / 2, r);
}

ll get(int a, int b, int k = 0, int l = 0, int r = -1){
    if(r < 0) r = n2;

    eval(k, l, r);
    if(b <= l || r <= a) return 0;
    if(a <= l && r <= b) return node[k];
    ll vl = get(a, b, k * 2 + 1, l, (l + r) / 2);
    ll vr = get(a, b, k * 2 + 2, (l + r) / 2, r);
    return vl + vr;
}

void init(){
    cin >> n >> q;
    n2 = 1;
    while(n2 < n) n2 <<= 1;
    node.resize(n2 * 2);
    lazy.resize(n2 * 2);
}

void solve(){
    for(int i= 0; i < q; i++){
        int query;
        cin >> query;
        if(query){
            int g;
            cin >> g;
            cout << get(g - 1, g) << endl;
        }else{
            int s, t, x;
            cin >> s >> t >> x;
            add(s - 1, t, x);
        }
    }
}

int main(){
    init();
    solve();
    return 0;
}