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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        B solver = new B();
        solver.solve(1, in, out);
        out.close();
    }

    static class B {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            // 入力
            int n = Integer.parseInt(in.next());
            int m = Integer.parseInt(in.next());
            int ab[][] = new int[n][2];
            for (int i = 0; i < n; i++) {
                ab[i][0] = Integer.parseInt(in.next());
                ab[i][1] = Integer.parseInt(in.next());
            }
            int cd[][] = new int[m][2];
            for (int i = 0; i < m; i++) {
                cd[i][0] = Integer.parseInt(in.next());
                cd[i][1] = Integer.parseInt(in.next());
            }
            int dist[][] = new int[n][m];

            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    dist[i][j] = Math.abs(ab[i][0] - cd[j][0]) + Math.abs(ab[i][1] - cd[j][1]);
                }
            }

            // 出力
            int ans = 0;
            int mindist = Integer.MAX_VALUE;
            for (int i = 0; i < n; i++) {
                ans = 0;
                for (int j = 0; j < m; j++) {
                    if (mindist > dist[i][j]) {
                        mindist = dist[i][j];
                        ans = j + 1;
                    }
                }
                out.println(ans);
                mindist = Integer.MAX_VALUE;
            }

        }

    }
}