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)

mod io{
    use std::io::*;
    pub fn read<T: ::std::str::FromStr>()->T{
        let stdin=stdin();
        let ret=stdin.lock().bytes().map(|c|c.unwrap()as char)
        .skip_while(|c|c.is_whitespace())
        .take_while(|c|!c.is_whitespace())
        .collect::<String>().parse().ok().unwrap();
        ret
    }
}
use io::read;
fn main(){
    let n:usize=read();
    let m:usize=read();
    let a=(0..n).map(|_|(read::<isize>(),read::<isize>())).collect::<Vec<(isize,isize)>>();
    let b=(0..m).map(|_|(read::<isize>(),read::<isize>())).collect::<Vec<(isize,isize)>>();
    for (x,y)in a{
        let mut s=1e9 as isize;
        let mut j=0 as usize;
        for i in 0..m{
            let now=(x-b[i].0).abs()+(y-b[i].1).abs();
            if now<s{s=now;j=i+1}
        }
        println!("{}",j);
    }
}