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: Wrong Answer)

import java.util.ArrayList;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		ArrayList<Integer> num = new ArrayList<Integer>();
		ArrayList<String> param = new ArrayList<String>();
		int system = 0; // 文字の空白区切り 0:ON 1:OFF

		while (sc.hasNext()) {
			if (sc.hasNextInt()) {
				num.add(sc.nextInt());
			} else {
				if (system == 0)
					param.add(sc.next());
				if (system == 1)
					param.add(sc.nextLine());
			}
		}
		Method(num, param);
	}

	static void Method(ArrayList<Integer> num, ArrayList<String> param) {
		int[] n = new int[num.size()];
		for (int i = 0; i < num.size(); i++) {
			n[i] = num.get(i);
		}
		String[] p = new String[param.size()];
		for (int i = 0; i < param.size(); i++) {
			p[i] = param.get(i);
		}
		////////////////////////////////////////////////////////////////////
		int[] kekka = new int[num.size()];
		long min = 999999999999999999l;
		int ansewr = 0;

		for (int i = 0; i < n[0] * 2; i += 2) {
			for (int j = 0; j < n[1]; j++) {
				kekka[j] = Math.abs(n[i + 2] - n[j * 2 + 2 + n[0] * 2]) + Math.abs(n[i + 3] - n[j * 2 + 3 + n[0] * 2]);
				//System.out.println(min + " " + kekka[j]);
				//if(kekka[i] > 100000000) while(true);
				System.out.println(kekka[j]);
				if (min > kekka[j]) {
					min = kekka[j];
					ansewr = j + 1;
				}
			}
			min = 999999999999999999l;
			System.out.println(ansewr);
		}

	}
}