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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <queue>
#include <utility>//pair
#include <cmath>//abs()
using namespace std;
#define MAX 200000
#define rep(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define in(n) scanf("%d",&n);
#define out(n) printf("%d\n",n);
#define in2(n,k) scanf("%d%d",&n,&k);
#define out2(n,k) printf("%d%d\n",n,k);
#define in3(n,k,s) scanf("%d%d%d",&n,&k,&s);
#define out3(n,k,s) printf("%d%d%d\n",n,k,s);

int main(){
    int n,m,dis;
    typedef pair<int,int>p;
    p a[50],b[50];
    in2(n,m);
    rep(i,n)in2(a[i].first,a[i].second);
    rep(i,m)in2(b[i].first,b[i].second);
    rep(i,n){
        int c=0;
        rep(j,m){
            int min=100000000*1.4;
            dis = abs(a[i].first-b[j].first)+abs(a[i].second-b[j].second);
            if(min>dis)min=dis,c=j;
        }
        out(c+1);
    }
    return 0;
}