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 <algorithm>
#include <chrono>
#include <iostream>
#include <vector>

using namespace std;
using namespace std::chrono;

int main() {
  auto start = high_resolution_clock::now();

  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);

  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> vs(n);
  for (int i = 0; i < n; ++i)
    cin >> vs[i].first >> vs[i].second;
  vector<pair<int, int>> vc(m);
  for (int i = 0; i < m; ++i)
    cin >> vc[i].first >> vc[i].second;

  for (int i = 0; i < n; ++i) {
    int ans = 0;
    for (int j = 1; j < m; ++j) {
      int dist = abs(vs[i].first - vc[ans].first) + abs(vs[i].second - vc[ans].second);
      int new_dist = abs(vs[i].first - vc[j].first) + abs(vs[i].second - vc[j].second);
      if (new_dist < dist)
        ans = j;
    }
    cout << ans + 1 << endl;
  }

  cerr << "Total execution time : " << duration_cast<milliseconds>(high_resolution_clock::now() - start).count() << " ms" << endl;

  return 0;
}