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 Scala (2.11.7) (judge verdict: Wrong Answer)

object Main{
  def main(args:Array[String]) = {
    val sc = new java.util.Scanner(System.in)
    val n = sc.nextInt
    val p = sc.nextInt

    val x_ = Array.ofDim[BigInt](n)
    val y_ = Array.ofDim[BigInt](n)
    for(i <- 0 to n-1){
      x_(i) = sc.nextInt
      y_(i) = sc.nextInt
    }

    val x = x_.sorted
    val y = y_.sorted

    //println(x.deep)
    //println(y.deep)

    var ans : BigInt = (x(n-1)-x(0)) * (y(n-1)-y(0))
    for(i <- 0 to n-1){
      for(j <- i+1 to n-1){
        for(k <- 0 to n-1){
          for(l <- k+1 to n-1){
            var cnt = 0
            for(m <- 0 to n-1){
              if(x(i) <= x(m) && x(m) <= x(j) && y(k) <= y(m) && y(m) <= y(l)) cnt = cnt + 1
            }
            //val t = (x(j)-x(i)) * (y(l)-y(k))
            //println(cnt,t)
            if(cnt >= p){
              ans = ans.min((x(j)-x(i))*(y(l)-y(k)))
            }
          }
        }
      }
    }
    println(ans)
  }
}