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 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void)
{
	int n,m;
	int i;
	int p;
	int k;
	int a[50];
	int b[50];
	int c[50];
	int d[50];
	int aa[50];
	int num[50];
	int temp;
	
	scanf("%d%d",&n,&m);
	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++){
		for(p=0;p<m;p++){
		aa[p]=(((a[i])-(c[p]))+((b[i])-(d[p])));
		num[p]=p+1;
		}
		k=i;
		
	
	for(i=0;i<m;i++){
		
		for(p=m-1;p>i;p--){
			
			if(aa[p]<aa[p-1]){
				temp=aa[p-1];
				aa[p-1]=aa[p];
				aa[p]=temp;
				temp=num[p-1];
				num[p-1]=num[p];
				num[p]=temp;
				
				
				
				
				
				
				
				
			}
			
			
			
			
			
			
			
			
			
			
			
			
		}
		
		
		
		
		
		
	}
//printf("\n");
	/*for(i=0;i<m;i++){
		
		printf("%d %d\n",aa[i],num[i]);
		
	}*/
	//printf("\n");
	i=k;
	printf("%d\n",num[m-1]);
	}
	
	return 0;
}