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>

using namespace std;

const int N=1e5+10;

typedef vector<int> vec;

class RUQ
{
public:
    vec data;
    vec lazy;
    int n;

    RUQ(int _n)
    {
        data.resize(4*_n);
        lazy.resize(4*_n);
        for(n=1;n<_n;n<<=1);
        fill(data.begin(),data.end(),0);
        fill(lazy.begin(),lazy.end(),0);
    }

    void Down(int p,int l,int r)
    {
        if(l+1>=r)
        {
            data[p]+=lazy[p];
        }else
        {
            lazy[p*2+1]+=lazy[p];
            lazy[p*2+2]+=lazy[p];
        }
        lazy[p]=0;
    }

    void update(int p,int l,int r,int x,int a,int b)
    {
        if(a>=r||b<=l) return;
        if(a<=l&&r<=b)
        {
            lazy[p]+=x;
            return;
        }
        Down(p,l,r);
        int mid=(l+r)/2;
        update(p*2+1,l,mid,x,a,b);
        update(p*2+2,mid,r,x,a,b);
    }

    int query(int p,int l,int r,int a,int b)
    {
        if(a>=r||b<=l) return -1;
        Down(p,l,r);
        if(a<=l&&r<=b) return data[p];
        int mid=(l+r)/2;
        if(a<mid)
        {
            return query(p*2+1,l,mid,a,b);
        }else
        {
            return query(p*2+2,mid,r,a,b);
        }
    }
};

int main()
{
    int n,q;

    while(scanf("%d %d",&n,&q)!=EOF)
    {

        RUQ ruq(n);
        while(q--)
        {
            int op,a,b,c;
            scanf("%d",&op);
            if(op)
            {
                scanf("%d",&a);
                printf("%d\n",ruq.query(0,0,ruq.n,a-1,a));
            }else
            {
                scanf("%d %d %d",&a,&b,&c);
                ruq.update(0,0,ruq.n,c,a-1,b);
            }
        }
    }

    return 0;
}