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::env;
use std::io::*;
use std::str::*;
use std::mem;
use std::collections::HashMap;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let s = stdin.bytes().map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.parse::<T>().ok().unwrap()
}
fn read_char() -> char {
    let c = read::<String>();
    let c = c.chars().collect::<Vec<char>>();
    c[0]
}
fn read_String_Vec() -> Vec<char> {
    let S = read::<String>();
    S.chars().collect::<Vec<char>>()
}

fn vecchar_to_String(Vs: Vec<char>) -> String {
    let mut S = String::new();
    for i in 0..Vs.len() {
        S.push(Vs[i]);
    }
    S
}

/*
A : 65
Z : 90
a : 97
z : 122
*/

/*
let N = read::<i32>();
let S = read::<String>();
let S = read_String_Vec();
let C = read_char();
*/

/*
 .to_string() : * -> String
 &*S          : String -> &str
*/

fn minidist((x, y): (i32, i32), chkpnt: & Vec<(i32, i32)>) -> i32 {
    let len = chkpnt.len();
    let mut minimumpoint: i32 = -1;
    let mut minimumdistance = <i32>::max_value();
    for i in 0..len {
        let (px, py) = chkpnt[i];
        let dist = (x - px).abs() + (y - py).abs();
        if dist < minimumdistance {
            minimumdistance = dist;
            minimumpoint = i as i32;
        }
    }
    minimumpoint+1
}

fn main() {
    // main の引数
    /*
        引数なしのときargs.len() = 1;
    */
    let args: Vec<String> = env::args().collect();

    let N = read::<usize>();
    let M = read::<usize>();
    let mut student: Vec<(i32, i32)> = Vec::new();
    let mut checkpnt: Vec<(i32, i32)> = Vec::new();

    for n in 0..N {
        let x = read::<i32>();
        let y = read::<i32>();
        student.push( (x, y) );
    }

    for m in 0..M {
        let x = read::<i32>();
        let y = read::<i32>();
        checkpnt.push( (x, y) );
    }
    for n in 0..N {
        let tmp = minidist(student[n], & checkpnt);
        println!("{}", tmp);
    }

}