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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
// warm-up
// Good riddance manhattan. No more manhattan new york california please.
public class Main {

	static class Point {
		int x=0, y=0;
		Point (int a, int b) {
			x=a; y=b;
		}
	}

	static void solve() {
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt(), m=sc.nextInt(), t=0, i=0;
		Point[] a = new Point[n];
		Point[] b = new Point[m];
		t=n;
		while (t-->0) a[i++] = new Point(sc.nextInt(), sc.nextInt()); 
		t=m; i=0;
		while (t-->0) b[i++] = new Point(sc.nextInt(), sc.nextInt()); 
		for (Point d : a) {
			int min=Integer.MAX_VALUE, j=0; i=1;
			for (Point c : b) {
				if ((Math.abs(d.x-c.x)+Math.abs(d.y-c.y))<min) {
					min = Math.abs(d.x-c.x)+Math.abs(d.y-c.y);
					j=i;
				}
				i++;
			}
			System.out.println(j);
		}
		sc.close();			
	}

	public static void main(String args[]) {
		solve();
	}

}