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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"strconv"
	"strings"
	"os"
)

func readLine(sc *bufio.Scanner) string {
	sc.Scan()
	return sc.Text()
}

func readInt(sc *bufio.Scanner) (res int) {
	res, _ =strconv.Atoi(readLine(sc))
	return
}

func readInts(sc *bufio.Scanner, n int) []int {
	ns := strings.Split(readLine(sc), " ")
	res := make([]int, n)

	for i, n := range ns {
		res[i], _ = strconv.Atoi(n)
	}
	return res
}

func readInt64s(sc *bufio.Scanner, n int) []int64 {
	ns := strings.Split(readLine(sc), " ")
	res := make([]int64, n)

	for i, n := range ns {
		res[i], _ = strconv.ParseInt(n, 10, 64)
	}
	return res
}

func readMatrix(sc *bufio.Scanner, rows, cols int) [][]int {
	res := make([][]int, rows)
	for i := 0; i < rows; i ++ {
		res[i] = readInts(sc, cols)
	}
	return res
}

func abs(i int) int {
	if i < 0 {
		return -i
	} else {
		return i
	}
}

func main() {
	sc := bufio.NewScanner(os.Stdin)

	xs := readInts(sc, 2)
	n := xs[0]
	m := xs[1]

	students := readMatrix(sc, n, 2)
	points := readMatrix(sc, m, 2)

	for _, s := range students {
		x := s[0]
		y := s[0]

		min_i := -1
		min := 1000000000
		for j := len(points) - 1; j >= 0; j-- {
			u := points[j][0]
			v := points[j][1]

			d := abs(u - x) + abs(v - y)
			if d <= min {
				min_i = j
			}
		}

		fmt.Println(min_i + 1)
	}
}