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: Accepted)

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <time.h>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <string.h>

#define sf scanf
#define pf printf
#define lf double
#define ll long long
#define p123 printf("123\n");
#define pn printf("\n");
#define pk printf(" ");
#define p(n) printf("%d",n);
#define pln(n) printf("%d\n",n);
#define s(n) scanf("%d",&n);
#define ss(n) scanf("%s",n);
#define ps(n) printf("%s",n);
#define sld(n) scanf("%lld",&n);
#define pld(n) printf("%lld",n);
#define slf(n) scanf("%lf",&n);
#define plf(n) printf("%lf",n);
#define sc(n) scanf("%c",&n);
#define pc(n) printf("%c",n);
#define gc getchar();
#define re(n,a) memset(n,a,sizeof(n));
#define len(a) strlen(a)
using namespace std;
struct A{
    ll x,y;
    bool flag = false;
}a[1000],b[1000];
int main() {
    ll n,m;
    sld(n) sld(m)
    for(ll i = 0; i < n; i ++){
        sld(a[i].x) sld(a[i].y)
    }
    for(ll i = 0; i < m; i ++){
        sld(b[i].x) sld(b[i].y)
    }
    for(ll i = 0; i < n; i ++){
        ll mini = 0;
        ll minn = 10000000000;
        for(ll j = 0; j < m; j ++){
            if(abs(a[i].x-b[j].x)+abs(a[i].y-b[j].y) < minn){
                mini = j;
                minn = abs(a[i].x-b[j].x)+abs(a[i].y-b[j].y);
            }
        }
        pld(mini +1ll) pn
        b[mini].flag = true;
    }
    return 0;
}