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)

use std::io::{BufRead, Write, stdin, stdout};

fn solve<R: BufRead, W: Write>(mut input: R, mut output: W) {
    let mut s = String::new();
    input.read_line(&mut s).unwrap();
    let mut split = s.split_whitespace();
    let n = split.next()
        .unwrap()
        .parse::<usize>()
        .unwrap();
    let m = split.next()
        .unwrap()
        .parse::<usize>()
        .unwrap();
    let ab = (0..n)
        .map(|_| {
            let mut s = String::new();
            input.read_line(&mut s).unwrap();
            let mut split = s.split_whitespace();
            let a = split.next()
                .unwrap()
                .parse::<i64>()
                .unwrap();
            let b = split.next()
                .unwrap()
                .parse::<i64>()
                .unwrap();
            (a, b)
        })
        .collect::<Vec<_>>();
    let cd = (0..m)
        .map(|_| {
            let mut s = String::new();
            input.read_line(&mut s).unwrap();
            let mut split = s.split_whitespace();
            let c = split.next()
                .unwrap()
                .parse::<i64>()
                .unwrap();
            let d = split.next()
                .unwrap()
                .parse::<i64>()
                .unwrap();
            (c, d)
        })
        .collect::<Vec<_>>();
    for (sx, sy) in ab {
        writeln!(output,
                 "{}",
                 cd.iter()
                     .map(|&(tx, ty)| (sx - tx).abs() + (sy - ty).abs())
                     .enumerate()
                     .min_by(|&(_, dist1), &(_, dist2)| dist1.cmp(&dist2))
                     .unwrap()
                     .0 + 1)
                .unwrap();
    }
}

fn main() {
    let stdin = stdin();
    let stdout = stdout();
    solve(stdin.lock(), stdout.lock());
}

#[cfg(test)]
mod tests {
    fn test(input: &str, output: &str) {
        let mut v = Vec::<u8>::new();
        ::solve(input.as_bytes(), &mut v);
        assert_eq!(String::from_utf8(v).unwrap(), output);
    }
    #[test]
    fn case1() {
        test("2 2
2 0
0 0
-1 0
1 0\n",
             "2
1\n");
    }
    #[test]
    fn case2() {
        test("3 4
10 10
-10 -10
3 3
1 2
2 3
3 5
3 5\n",
             "3
1
2\n");
    }
    #[test]
    fn case3() {
        test("5 5
-100000000 -100000000
-100000000 100000000
100000000 -100000000
100000000 100000000
0 0
0 0
100000000 100000000
100000000 -100000000
-100000000 100000000
-100000000 -100000000\n",
             "5
4
3
2
1\n");
    }
}