Problem:
Score : 200 points

Problem Statement

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.

Sample Input 1

2 2
2 0
0 0
-1 0
1 0

Sample Output 1

2
1

The Manhattan distance between the first student and each checkpoint is:

For checkpoint 1: |2-(-1)|+|0-0|=3

For checkpoint 2: |2-1|+|0-0|=1

The nearest checkpoint is checkpoint 2. Thus, the first line in the output should contain 2.

The Manhattan distance between the second student and each checkpoint is:

For checkpoint 1: |0-(-1)|+|0-0|=1

For checkpoint 2: |0-1|+|0-0|=1

When there are multiple nearest checkpoints, the student will go to the checkpoint with the smallest index. Thus, the second line in the output should contain 1.

Sample Input 2

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

Sample Output 2

3
1
2

There can be multiple checkpoints at the same coordinates.

Sample Input 3

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

Sample Output 3

5
4
3
2
1

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
	    int N = sc.nextInt();
	    int M = sc.nextInt();
	    
	    int [] A = new int [N];
	    int [] B = new int [N];
	    
	    for (int i=0; i<N; i++){
	    	A[i]=sc.nextInt();
	    	B[i]=sc.nextInt();
	    }
	    
	    int [] C = new int [M];
	    int [] D = new int [M];
	    
	    for (int i=0; i<M; i++){
	    	C[i]=sc.nextInt();
	    	D[i]=sc.nextInt();
	    }
	    
	    for (int i=0; i<N; i++){
	    	int a = A[i];
	    	int b = B[i];
	    	int c = C[0];
	    	int d = D[0];
	    	
	    	int min = Math.abs(a-c)+Math.abs(b-d);
	    	int index = 1;
	    	
	    	for (int j=1; j<M; j++){
	    		c = C[j];
	    		d = D[j];
	    		int temp = Math.abs(a-c)+Math.abs(b-d);
	    		
	    		if (temp<min){
	    			index=j+1;
	    			min=temp;
	    		}
	    			
	    	}
	    	System.out.println(index);
	    }
	    sc.close();
	    
	    

	}

}