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 <algorithm>

#define MAX 50
int N;                   //学生の数
int M;                   //チェックポイントの数
struct point{            //座標の構造体
  int x;                 //x座標
  int y;                 //y座標
};
point student[MAX];      //学生の位置座標
point check[MAX];        //チェックポイントの位置座標
unsigned long int s;          //マンハッタン距離
struct sum{
  unsigned long int s_min;    //最小マンハッタン距離
  int number;            //最小マンハッタン距離となるチェックポイント番号
}result[MAX];            //各学生の結果
int i,j;                 //カウント変数

int main()
{
  std::cin >> N >> M;
  for(i=0;i<N;i++){
    std::cin >> student[i].x >> student[i].y;
  }
  for(j=0;j<M;j++){
    std::cin >> check[j].x >> check[j].y;
  }

  for(i=0;i<N;i++){
    result[i].s_min = 100000000;          //初期化
    result[i].number = 0;                 //初期化
    for(j=0;j<M;j++){
      s = std::abs(student[i].x-check[j].x) + std::abs(student[i].y-check[j].y);
      if(s < result[i].s_min){
        result[i].s_min = s;
        result[i].number = j;
      }
    }
    std::cout << ++result[i].number << '\n';
  }

  return (0);
}