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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int a[55];
int b[55];
int c[55];
int d[55];

int max(int a,int b);

int main() {

	int N, M;

	scanf("%d %d",&N,&M);

	int i,j;

	for (i = 0; i < N; i++) {
		scanf("%d %d",&a[i],&b[i]);
	}
	for (i = 0; i < M; i++) {
		scanf("%d %d", &c[i], &d[i]);
	}

	for (i = 0; i < N; i++) {
		int ma = 1000000000;
		int ans = -1;
		for (j = 0; j < M; j++) {
			if (ma > max(a[i] - c[j], c[j] - a[i]) + max(b[i] - d[j], d[j] - b[i])) {
				ma = max(a[i] - c[j], c[j] - a[i]) + max(b[i] - d[j], d[j] - b[i]);
				ans = j + 1;
			}
		}
		printf("%d\n",ans);
	}

	return 0;
}
int max(int a, int b) {
	if (a >= b) { return a; }
	else { return b; }
}