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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for (int i=0; i<(int)(n); i++)
#define REP(i,m,n) for(ll i=(ll)(m);i<(ll)(n);i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
  int n, m; cin >> n >> m;
  const int INF = 100000008;
  vector<P> s(n), p(m), ans(n, make_pair(0,INF));
  rep(i,n) {
    int a, b; cin >> a >> b;
    s[i] = make_pair(a,b);
  }
  rep(i,m) {
    int c, d; cin >> c >> d;
    p[i] = make_pair(c,d);
  }
  rep(i,n) {
    ll x1 = s[i].first;
    ll y1 = s[i].second;
    rep(j,m) {
      ll x2 = p[j].first;
      ll y2 = p[j].second;
      ll d = abs(x1 - x2) + abs(y1 - y2);
      if (ans[i].second > d) {
        ans[i].first = j+1;
        ans[i].second = d;
      }
    }
  }
  rep(i,n) {
    cout << ans[i].first << endl;
  }

  return 0;
}