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: Accepted)

#include <bits/stdc++.h>
#define rep(i, f, n) for (ll i = (f); i < (ll)(n); i++)
#define repe(i, f, n) for (ll i = (f); i <= (ll)(n); i++)
using namespace std;
using ll = long long;
using pint = pair<int, int>;

const int INF = 100100100;

int main() {
  int N, M, a, b, c, d;
  cin >> N >> M;
  vector<pint> starts(N);
  rep(i, 0, N) {
    cin >> a >> b;
    starts[i] = pint(a, b);
  }
  vector<pint> goals(M);
  rep(i, 0, M) {
    cin >> c >> d;
    goals[i] = pint(c, d);
  }

  rep(i, 0, N) {
    int x1 = starts[i].first, y1 = starts[i].second;
    int ans = 1, mi = abs(x1 - goals[0].first) + abs(y1 - goals[0].second);
    rep(j, 1, M) {
      int x2 = goals[j].first, y2 = goals[j].second;
      int m = abs(x1 - x2) + abs(y1 - y2);
      if (mi > m) {
        mi = m;
        ans = j + 1;
      }
    }

    cout << ans << endl;
  }

  return 0;
}