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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
public class Main {

  public static void main(String[] args) { 
      Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      int k = sc.nextInt();
      long[][] cood = new long[n][2];
      for(int i = 0; i < n; i++) {
        cood[i][0] = sc.nextInt();
        cood[i][1] = sc.nextInt();
      }

      long maxArea = Long.MAX_VALUE;
      int cnt = 0;

      for(long[] a : cood) {
        for(long[] b : cood) {
          for(long[] c : cood) {
            for(long[] d: cood) {
              cnt = 0;
              long leftX = Math.min(Math.min(a[0], b[0]), Math.min(c[0], d[0])), 
                  rightX = Math.max(Math.max(a[0], b[0]), Math.max(c[0], d[0]));
              long downY = Math.min(Math.min(a[1], b[1]), Math.min(c[1], d[1])), 
                  upY = Math.max(Math.max(a[1], b[1]), Math.max(c[1], d[1]));
              for(long[] p : cood) {
                if(leftX <= p[0] && p[0] <= rightX &&
                  downY <= p[1] && p[1] <= upY
                ) cnt++;
              }
             // System.out.println(rightX + "," + leftX + " - " + upY + "," + downY);
              if(cnt >= k) {
                  
                maxArea = Math.min(maxArea, (rightX - leftX) * (upY - downY));
              }
            }
          }
        }
      }
      System.out.println(maxArea);
  }
}