#include<bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
#define int long long
using namespace std;
int a[55],b[55],c[55],d[55];
signed main(){
	int n,m;cin>>n>>m;
	rep(i,n)cin>>a[i]>>b[i];
	rep(i,m)cin>>c[i]>>d[i];
	
	rep(i,n){
		int dis=1145141919,num=-1;
		rep(j,m){
			int di=abs(a[i]-c[j])+abs(b[i]-d[j]);
			if(dis>di){
				dis=di;
				num=j;
				}
			}
			cout<<num+1<<endl;
		}
	}

### Prompt

Please create a solution in Cpp to the following problem:

There are N students and M checkpoints on the xy-plane.
The coordinates of the i-th student (1 \leq i \leq N) is (a_i,b_i), and the coordinates of the checkpoint numbered j (1 \leq j \leq M) is (c_j,d_j).
When the teacher gives a signal, each student has to go to the nearest checkpoint measured in Manhattan distance.
The Manhattan distance between two points (x_1,y_1) and (x_2,y_2) is |x_1-x_2|+|y_1-y_2|.
Here, |x| denotes the absolute value of x.
If there are multiple nearest checkpoints for a student, he/she will select the checkpoint with the smallest index.
Which checkpoint will each student go to?

Constraints

* 1 \leq N,M \leq 50
* -10^8 \leq a_i,b_i,c_j,d_j \leq 10^8
* All input values are integers.

Input

The input is given from Standard Input in the following format:


N M
a_1 b_1
:
a_N b_N
c_1 d_1
:
c_M d_M


Output

Print N lines.
The i-th line (1 \leq i \leq N) should contain the index of the checkpoint for the i-th student to go.

Examples

Input

2 2
2 0
0 0
-1 0
1 0


Output

2
1


Input

3 4
10 10
-10 -10
3 3
1 2
2 3
3 5
3 5


Output

3
1
2


Input

5 5
-100000000 -100000000
-100000000 100000000
100000000 -100000000
100000000 100000000
0 0
0 0
100000000 100000000
100000000 -100000000
-100000000 100000000
-100000000 -100000000


Output

5
4
3
2
1

### Response

```cpp
#include<bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
#define int long long
using namespace std;
int a[55],b[55],c[55],d[55];
signed main(){
	int n,m;cin>>n>>m;
	rep(i,n)cin>>a[i]>>b[i];
	rep(i,m)cin>>c[i]>>d[i];
	
	rep(i,n){
		int dis=1145141919,num=-1;
		rep(j,m){
			int di=abs(a[i]-c[j])+abs(b[i]-d[j]);
			if(dis>di){
				dis=di;
				num=j;
				}
			}
			cout<<num+1<<endl;
		}
	}

```