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++)
#define rrep(i,n) for(int i=1;i<=n;i++)
#define drep(i,n) for(int i=n;i>=0;i--)
#define INF 100000005
#define MAX 131072
#define ll long long
#define dmp make_pair
#define dpb push_back
#define P pair<int,int>
#define fi first
#define se second
using namespace std;
//__gcd(a,b), __builtin_popcount(a);
const int inf = 2147483647;

int n, dat[2*MAX];
void init(int m){
	n = 1;
	while(n < m)n *= 2;
	for(int i = 0;i < 2*n;i++)dat[i] = 0;
}

/*void change(int x, int y){
	x += n-1;
	dat[x] += y;
	while(x > 0){
		x = (x-1)/2;
		dat[x] += y;
	}
}

int query(int a, int b, int k, int l, int r){
	if(b <= l || r <= a)return 0;
	if(a <= l && r <= b)return dat[k];
	int vl = query(a, b, 2*k+1, l, (l+r)/2);
	int vr = query(a, b, 2*k+2, (l+r)/2, r);
	return vl+vr;
}*/

int add(int a, int b, int x,  int k, int l, int r){
	if(b <= l || r <= a)return 0;
	if(a <= l && r <= b){
		dat[k] += x;
		return 0;
	}
	add(a, b, x, k*2+1, l, (l+r)/2);
	add(a, b, x, k*2+2, (l+r)/2, r);
	return 0;
}

int get(int k){
	int ans = 0;
	k += n-1;
	ans += dat[k];
	while(k > 0){
		k = (k-1)/2;
		ans += dat[k];
	}
	return ans;
}

int main(){
	int m, q, c, s, t, x, ans;
	scanf("%d%d", &m, &q);
	init(m);
	while(q--){
		scanf("%d", &c);
		if(!c){
			scanf("%d%d%d", &s, &t, &x);
			add(s-1, t, x, 0, 0, n);
		}else{
			scanf("%d", &x);
			printf("%d\n", get(x-1));
			//printf("%d\n", get(x-1, y, 0, 0, n));
		}
	}
	//for(int i = 0;i < 2*n-1;i++)printf("%d ", dat[i]);
	return 0;
}