Problem:
Score : 400 points

Problem Statement

We have N points in a two-dimensional plane.

The coordinates of the i-th point (1 \leq i \leq N) are (x_i,y_i).

Let us consider a rectangle whose sides are parallel to the coordinate axes that contains K or more of the N points in its interior.

Here, points on the sides of the rectangle are considered to be in the interior.

Find the minimum possible area of such a rectangle.

Constraints

2 \leq K \leq N \leq 50

-10^9 \leq x_i,y_i \leq 10^9 (1 \leq i \leq N)

x_i≠x_j (1 \leq i<j \leq N)

y_i≠y_j (1 \leq i<j \leq N)

All input values are integers. (Added at 21:50 JST)

Input

Input is given from Standard Input in the following format:

N K
x_1 y_1
:
x_{N} y_{N}

Output

Print the minimum possible area of a rectangle that satisfies the condition.

Sample Input 1

4 4
1 4
3 3
6 2
8 1

Sample Output 1

21

One rectangle that satisfies the condition with the minimum possible area has the following vertices: (1,1), (8,1), (1,4) and (8,4).

Its area is (8-1) × (4-1) = 21.

Sample Input 2

4 2
0 0
1 1
2 2
3 3

Sample Output 2

1

Sample Input 3

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999

Sample Output 3

3999999996000000001

Watch out for integer overflows.

Solution in Go (1.6) (judge verdict: Accepted)

package main

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

var sc = bufio.NewScanner(os.Stdin)

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

func main() {
	buf := make([]byte, 0)
	sc.Buffer(buf, 1000000007)
	sc.Split(bufio.ScanWords)
	n, _ := strconv.Atoi(Scanner())
	k, _ := strconv.Atoi(Scanner())
	xs := make([]int, n)
	ys := make([]int, n)
	for i := 0; i < n; i++ {
		xs[i], _ = strconv.Atoi(Scanner())
		ys[i], _ = strconv.Atoi(Scanner())
	}
	sorted_xs := make([]int, n)
	sorted_ys := make([]int, n)
	copy(sorted_xs, xs)
	copy(sorted_ys, ys)
	sort.Ints(sorted_xs)
	sort.Ints(sorted_ys)
	ans := 4000000000000000000
	for i := 0; i < n; i++ {
		for j := i + 1; j < n; j++ {
			for p := 0; p < n; p++ {
				for l := p + 1; l < n; l++ {
					c := 0
					for m := 0; m < n; m++ {
						if xs[m] >= sorted_xs[i] && xs[m] <= sorted_xs[j] && ys[m] >= sorted_ys[p] && ys[m] <= sorted_ys[l] {
							c++
						}
					}
					if c >= k {
						s := (sorted_xs[j] - sorted_xs[i]) * (sorted_ys[l] - sorted_ys[p])
						if s == 0 {
							ans = smaller(ans, smaller((sorted_xs[j]-sorted_xs[i]), (sorted_ys[l]-sorted_ys[p])))
						} else {
							ans = smaller(ans, s)
						}
					}
				}
			}
		}
	}
	fmt.Println(larger(ans, 1))
}
func smaller(x, y int) int {
	if x < y {
		return x
	} else {
		return y
	}
}
func larger(x, y int) int {
	if x > y {
		return x
	} else {
		return y
	}
}