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)]

use std::cmp::{max, min};
use std::io::stdin;
use std::str::FromStr;

fn input<T: FromStr>() -> Result<T, T::Err> {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim().parse()
}

fn input_vec<T: FromStr>() -> Result<Vec<T>, T::Err> {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim().split_whitespace().map(|x| x.parse()).collect()
}

fn main() {
    let (n, m): (usize, usize) = {
        let tmp = input_vec().unwrap();
        (tmp[0], tmp[1])
    };

    let mut stu: Vec<(i64, i64)> = Vec::with_capacity(n);
    for _i in 0..n {
        let tmp = {
            let t = input_vec().unwrap();
            (t[0], t[1])
        };
        stu.push(tmp);
    }

    let mut chk: Vec<(i64, i64)> = Vec::with_capacity(m);
    for _i in 0..m {
        let tmp = {
            let t = input_vec().unwrap();
            (t[0], t[1])
        };
        chk.push(tmp);
    }

    for &(a, b) in stu.iter() {
        let mut closest = m + 1;
        let mut ord = std::usize::MAX;

        for (i, &(c, d)) in chk.iter().enumerate().rev() {
            if ord
                >= ((a as i64) - (c as i64)).abs() as usize
                    + ((b as i64) - (d as i64)).abs() as usize
            {
                closest = i;
                ord = ((a as i64) - (c as i64)).abs() as usize
                    + ((b as i64) - (d as i64)).abs() as usize;
            }
        }

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