#include <iostream>
#include <algorithm>
using namespace std;

int main(){
  long long i,j,n,m;
  cin >> n >> m;
  long long a[n],b[n],c[m],d[m];
  for(i=0;i<n;i++)cin >> a[i] >> b[i];
  for(i=0;i<m;i++)cin >> c[i] >> d[i];
  for(i=0;i<n;i++){
    long long ret = 3e8;
    long long ans = -1;
    for(j=0;j<m;j++){
      if(ret>(abs(a[i]-c[j])+abs(b[i]-d[j]))){
        ret = abs(a[i]-c[j])+abs(b[i]-d[j]);
        ans = j+1;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 <iostream>
#include <algorithm>
using namespace std;

int main(){
  long long i,j,n,m;
  cin >> n >> m;
  long long a[n],b[n],c[m],d[m];
  for(i=0;i<n;i++)cin >> a[i] >> b[i];
  for(i=0;i<m;i++)cin >> c[i] >> d[i];
  for(i=0;i<n;i++){
    long long ret = 3e8;
    long long ans = -1;
    for(j=0;j<m;j++){
      if(ret>(abs(a[i]-c[j])+abs(b[i]-d[j]))){
        ret = abs(a[i]-c[j])+abs(b[i]-d[j]);
        ans = j+1;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```