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>
int main(void){
    int i, k, temp, N, M;
    scanf("%d %d", &N, &M);
    int n[N][2];
    int m[M][2];
    int answer[N][2];
    for(i = 0; i < N; i++) {
        scanf("%d %d", &n[i][0], &n[i][1]);
    }
    for(i = 0; i < M; i++) {
        scanf("%d %d", &m[i][0], &m[i][1]);
    }
    
    for(i = 0; i < N; i++) {
        answer[i][1] = 100000000;
    }
    for(i = 0; i < N; i++) {
        for(k = 0; k < M; k++) {
            if(n[i][0] > m[k][0]) {
               temp = n[i][0] - m[k][0];
            }
            else {
               temp = m[k][0] - n[i][0];
            }
            if(n[i][1] > m[k][1]) {
               temp += n[i][1] - m[k][1];
            }
            else {
               temp += m[k][1] - n[i][1];
            }
            if (answer[i][1] > temp) {
                answer[i][0] = k + 1;
                answer[i][1] = temp;
            }
        }
    }
    for(i = 0; i < N; i++) {
        printf("%d\n", answer[i][0]);
    }
}