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: Wrong Answer)

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

#define GYOU_MAX 256
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define s_pri(str) printf("%s\n",str);
#define PRINT(str) printf(#str "\n")
#define v_pri(val) printf("%d\n",val);
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define pi acos(-1.0)
#define sca(x) scanf("%d",&x);

typedef long long int ll;

/*
puts("aa");
//*/



int main(void) {

	int N, M;
	sca(N); sca(M);
	int a[50], b[50], c[50], d[50];
	rep(i, N) {
		sca(a[i]);
		sca(b[i]);
	}
	rep(i, M) {
		sca(c[i]);
		sca(d[i]);
	}

	int kyori;
	int min = 200000000;
	int x;
	rep(i, N) {
		rep(j, N) {
			kyori = abs(c[j] - a[i]) + abs(d[j] - b[i]);
			if (kyori < min) {
				min = kyori;
				x = j;
			}
			if (j == N - 1) {
				min = 200000000;
			}
			//printf("%d\n", kyori);
		}
		printf("%d\n", x+1);
	}
	
	
	return 0;
}