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<cstdio>
#include<algorithm>
#define N 100000
#define MAX (1<<31)-1
using namespace std;
long long A[N+1], T[4*N+1], lazy[4*N+1];

void build(int l, int r, int k) {
    if (l == r) {
        T[k] = A[l];
        return;
    }
    int mid = (l + r) / 2;
    build(l, mid, k * 2);
    build(mid + 1, r, k * 2 + 1);
    T[k] = T[2 * k] + T[2 * k + 1];
}
void pushdown(int k, int l, int r, int mid) {
    if (lazy[k]) {
        lazy[2*k] += lazy[k];
        lazy[2*k+1] += lazy[k];
        T[2*k] += lazy[k]*(mid-l+1);
        T[2*k+1] += lazy[k]*(r-mid);
        lazy[k] = 0;
    }
}
void updata(int L, int R, int data ,int l, int r ,int k) {
    if (L <= l && r <= R) {
        T[k] += data*(r-l+1);
        lazy[k] += data;
        return;
    }
    int mid = (l + r) / 2;
    pushdown(k, l, r, mid);
    if (mid >= L) updata(L, R, data, l, mid, 2*k);
    if (mid < R) updata(L, R, data, mid+1, r, 2*k+1);
    T[k] = T[2 * k] + T[2 * k + 1];
}
long long find(int L, int R, int l, int r, int k) {
    if (L <= l && r <= R) return T[k];
    int mid = (l + r) / 2;
    pushdown(k, l, r, mid);
    long long ans = 0;
    if (mid >= L) ans = ans + find(L, R, l, mid, k * 2);
    if (mid < R) ans = ans + find(L, R, mid + 1, r, k * 2 + 1);
    return ans;
}
int main() {
    int n, q;
    scanf("%d%d", &n, &q);
    for (int i = 0; i <= n; i++)
        A[i] = 0;
    for (int i = 0; i <= 4*n; i++)
        lazy[i] = 0;
    build(1, n, 1);
    int com, s, t, x, y;
    while (q--) {
        scanf("%d", &com);
        if (com == 0) {
            scanf("%d%d%d", &s, &t, &x);
            updata(s, t, x, 1, n, 1);
        }
        else {
            scanf("%d", &s);
            printf("%ld\n", find(s, s, 1, n, 1));
        }
    }
    return 0;
}