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 (1.15.1) (judge verdict: Accepted)

#[allow(unused_imports)]
use std::io;
use std::io::stdin;

fn read_line_as_vec() -> Vec<String> {
    let mut s = String::new();
    stdin().read_line(&mut s).expect("");
    s.split(' ').map(|x| x.to_string()).collect::<Vec<String>>()
}

fn main() {
    let nm = read_line_as_vec();
    let n: i32 = nm[0].trim().parse().expect("n");
    let m: i32 = nm[1].trim().parse().expect("m");
    let mut students = Vec::<(i32,i32)>::new();
    for _ in 0..n {
        let ab = read_line_as_vec();
        let a: i32 = ab[0].trim().parse().expect("");
        let b: i32 = ab[1].trim().parse().expect("");
        students.push((a, b));
    }
    let students = students;
    let mut points = Vec::<(i32,i32)>::new();
    for _ in 0..m {
        let cd = read_line_as_vec();
        let c: i32 = cd[0].trim().parse().expect("");
        let d: i32 = cd[1].trim().parse().expect("");
        points.push((c, d));
    }
    let points = points;
    for (a,b) in students {
        let mut argmin_cp = 0;
        let mut min_dist: i32 = 1_000_000_000;
        for (i,&(c,d)) in points.iter().enumerate() {
            let dist: i32 = (c-a).abs() + (d-b).abs();
            if dist < min_dist {
                min_dist = dist;
                argmin_cp = i + 1;
            }
        }
        println!("{}", argmin_cp);
    }
}