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
import java.util.Scanner;

public class Main {
	public static void main(String[] args){
		
		Scanner scan = new Scanner(System.in);
		
		int N = scan.nextInt();
		int M = scan.nextInt();
		int[]DN = new int[N];//一人の生徒の最小のマンハッタン距離
		int[]MN = new int[N];//一人の生徒の最小のマンハッタン距離の番号
		
		int[] a = new int[N];
		int[] b = new int[N];
		for(int i = 0;i<N;i++){
			a[i]=scan.nextInt();
			b[i]=scan.nextInt();
		}
		
		int[] c = new int[M];
		int[] d = new int[M];
		for(int i = 0;i<M;i++){
			c[i]=scan.nextInt();
			d[i]=scan.nextInt();
		}
		
		
		for(int i = 0;i<N;i++){//各生徒分繰り返す
			for(int j = 0;j<M;j++){//チェックポイントM個比較
				if(j==0){
					DN[i]=(Math.abs(a[i]-c[j])+Math.abs(b[i]-d[j]));
					MN[i]=j+1;
				}else{
					if(DN[i]>Math.abs(a[i]-c[j])+Math.abs(b[i]-d[j])){
						DN[i]=Math.abs(a[i]-c[j])+Math.abs(b[i]-d[j]);
						MN[i]=j+1;//ここで配列の〇番目を一番目似合わせる
					}
				}
				
			}
		}
		
		
		
		
		
		
		for(int i =0;i<N;i++){//最後に生徒分の最小のマンハッタン距離だったチェックポイントを示す
			
			//System.out.println(DN[i]);
			System.out.println(MN[i]);
		}
		
		
		
		
		
		
		
		
	}

}
