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 (Clang 10.0.0) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int closest(int a, int b, int *c, int *d, int M) {
	int i;
	int x, y;
	int ol_dist, new_dist;
	int point;
	for(i = 0; i < M; i += 1) {
		x = abs(a - c[i]);
		y = abs(b - d[i]);
		if(i == 0) {
			ol_dist = x + y;
			point = 0;
		} else {
			new_dist = x + y;
			if(new_dist < ol_dist) {
				ol_dist = new_dist;
				point = i;
			}
		}
	}
	return point;
}

int main(void) {
	int i;
	int N, M;
	int *a, *b, *c, *d;
	char buf[128], *ch;

	fgets(buf, sizeof(buf), stdin);
	ch = strtok(buf, " ");
	N = atoi(ch);
	ch = strtok(NULL, " ");
	M = atoi(ch);

	a = (int *)calloc(N, sizeof(int));
	b = (int *)calloc(N, sizeof(int));
	c = (int *)calloc(M, sizeof(int));
	d = (int *)calloc(M, sizeof(int));

	for(i = 0; i < N; i += 1) {
		fgets(buf, sizeof(buf), stdin);
		ch = strtok(buf, " ");
		a[i] = atoi(ch);
		ch = strtok(NULL, " ");
		b[i] = atoi(ch);
	}
	for(i = 0; i < M; i += 1) {
		fgets(buf, sizeof(buf), stdin);
		ch = strtok(buf, " ");
		c[i] = atoi(ch);
		ch = strtok(NULL, " ");
		d[i] = atoi(ch);
	}

	for(i = 0; i < N; i += 1) {
		printf("%d\n", 1 + closest(a[i], b[i], c, d, M));
	}

	free(a);
	free(b);
	free(c);
	free(d);
	return 0;
}