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 <vector>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <stack>
#include <math.h>
#include <map>
#define Z class
#define ln cout<<'\n'
#define ll long long
#define rep(i, n) for(int i = 0; i < (n); ++i)
//ASCII a=97, A=65
// int H[N]; rep(i,N) scanf("%d", &H[i]);
// int max_x = *std::max_element(x.begin(), x.end());
// int min_y = *std::min_element(y.begin(), y.end());
using namespace std;
template<class T> inline bool chmax(T& a, T b){ if(a < b) { a=b; return 1; }return 0;}
template<class T> inline bool chmin(T& a, T b){ if(a > b) { a=b; return 1; }return 0;}
template<Z A>void pr(A a){cout<<a;ln;}
template<Z A,Z B>void pr(A a,B b){cout<<a<<' ';pr(b);}
template<Z A,Z B,Z C>void pr(A a,B b,C c){cout<<a<<' ';pr(b,c);}
template<Z A,Z B,Z C,Z D>void pr(A a,B b,C c,D d){cout<<a<<' ';pr(b,c,d);}
template<Z A>void PR(A a,ll n){rep(i,n){if(i)cout<<' ';cout<<a[i];}ln;}

int GCD(int a, int b) { return b ? GCD(b, a%b) : a; }
const long long INF = 1LL << 60;

int main(){
    int N,M; cin >> N >> M;
    long a[60]={0}; long b[60]={0}; rep(i,N) cin >> a[i] >> b[i];
    long c[60]={0}; long d[60]={0}; rep(i,M) cin >> c[i] >> d[i];
    
    rep(i,N){
        long res = 100000100;
        int check_point = 1;
        rep(j,M){
            long len = abs(a[i]-c[j]) + abs(b[i]-d[j]);
            if(res>len){
                res = len;
                check_point = j+1;
            }
        }
        pr(check_point);
    }
}