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 Rust (1.42.0) (judge verdict: Accepted)

use std::io::stdin;

fn main() {
    let mut is = String::new();
    stdin().read_line(&mut is).ok();
    let mut itr = is.split_whitespace().map(|e| e.parse::<i64>().unwrap());
    let n = itr.next().unwrap();
    let m = itr.next().unwrap();
    let mut x = vec![];
    let mut y = vec![];

    for _ in 0..n {
        let mut is = String::new();
        stdin().read_line(&mut is).ok();
        let mut itr = is.split_whitespace().map(|e| e.parse::<i64>().unwrap());
        let a = itr.next().unwrap();
        let b = itr.next().unwrap();
        x.push((a, b));
    }

    for _ in 0..m {
        let mut is = String::new();
        stdin().read_line(&mut is).ok();
        let mut itr = is.split_whitespace().map(|e| e.parse::<i64>().unwrap());
        let a = itr.next().unwrap();
        let b = itr.next().unwrap();
        y.push((a, b));
    }

    for (a, b) in x {
        let mut h = 1 << 30;
        let mut ans = 0;
        for (i, &(c, d)) in y.iter().enumerate() {
            let t = (a - c).abs() + (b - d).abs();
            if t < h {
                h = t;
                ans = i;
            }
        }

        println!("{}", ans + 1);
    }
}