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

class Point {
    int x;
    int y;

    public Point(final int x, final int y) {
        this.x = x;
        this.y = y;
    }
}

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        solve(in);
    }

    private static void build(final Scanner in, List<Point> coordinates, int n) {
        int x, y;
        for (int i = 0; i < n; i++) {
            x = in.nextInt();
            y = in.nextInt();
            coordinates.add(new Point(x, y));
        }
    }

    private static void solve(final Scanner in) {
        int n = in.nextInt();
        int m = in.nextInt();
        List<Point> start = new ArrayList<>();
        List<Point> end = new ArrayList<>();
        build(in, start, n);
        build(in, end, m);
        checkPoint(start, end);
    }

    private static void checkPoint(final List<Point> start, final List<Point> end) {
        for (final Point point : start) {
            findMinimum(point, end);
        }
    }

    private static int getDis(Point p1, Point p2) {
        return Math.abs(p1.x - p2.x) + Math.abs(p1.y - p2.y);
    }

    private static void findMinimum(final Point point, final List<Point> end) {
        int minDis = Integer.MAX_VALUE, currDis, idx = 0;
        for (int i = 0; i < end.size(); i++) {
            Point currPoint = end.get(i);
            currDis = getDis(point, currPoint);
            if (currDis < minDis) {
                minDis = currDis;
                idx = i;
            }
        }
        System.out.println(idx + 1);
    }
}