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;
//BEGIN CUT HERE
template <typename T,typename E>
struct SegmentTree{
  typedef function<T(T,E)> G;
  typedef function<T(E,E)> H;
  int n;
  G g;
  H h;
  T d1;
  E d0;
  vector<T> dat;
  vector<E> laz;
  SegmentTree(){};
  SegmentTree(int n_,G g,H h,T d1,E d0,
	      vector<T> v=vector<T>()):
    g(g),h(h),d1(d1),d0(d0){
    init(n_);
    if(n_==(int)v.size()) build(n_,v);
  }
  void init(int n_){
    n=1;
    while(n<n_) n*=2;
    dat.clear();
    dat.resize(n,d1);
    laz.clear();
    laz.resize(2*n-1,d0);
  }
  void build(int n_, vector<T> v){
    for(int i=0;i<n_;i++) dat[i]=v[i];
  }
  void update(int a,int b,E x,int k,int l,int r){
    if(r<=a||b<=l) return;
    if(a<=l&&r<=b){
      laz[k]=h(laz[k],x);
      return;
    }
    update(a,b,x,k*2+1,l,(l+r)/2);
    update(a,b,x,k*2+2,(l+r)/2,r);
  }
  void update(int a,int b,E x){
    update(a,b,x,0,0,n);
  }
  T query(int k){
    T c=dat[k];
    k+=n-1;
    E x=laz[k];
    while(k>0){
      k=(k-1)/2;
      x=h(x,laz[k]);
    }
    return g(c,x);
  }
};
//END CUT HERE
signed main(){
  cin.tie(0);
  ios::sync_with_stdio(0);
  int n,q;
  cin>>n>>q;
  SegmentTree<int,int> raq(n,
			   [](int a,int b){return a+b;},
			   [](int a,int b){return a+b;},
			   0,0);
  for(int i=0;i<q;i++){
    int c,s,t,x;
    cin>>c;
    if(c){
      cin>>x;
      cout<<raq.query(x-1)<<endl;
    }else{
      cin>>s>>t>>x;
      raq.update(s-1,t,x);
    }
  }
  return 0;
}

/*
  verified on 2017/11/05
  http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_2_E&lang=jp
*/