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 C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

long long int min(long long int x,long long int y){
  if(x<y) return x;
  else return y;
}

long long int max(long long int x,long long int y){
  if(x>y) return x;
  else return y;
}

int main(){
  int n,k,i,j,l,m,o,cnt=0;
  long long int x[50],y[50],xmin,xmax,ymin,ymax,s;
  s=4000000000000000000;
  scanf("%d %d",&n,&k);
  for(i=0;i<n;i++){
    scanf("%ld %ld",&x[i],&y[i]);
  }
  for(i=0;i<n;i++){
    for(j=0;j<n;j++){
      if(i!=j){
        xmin=min(x[i],x[j]);
        xmax=max(x[i],x[j]);
        for(l=0;l<n;l++){
          for(m=0;m<n;m++){
            if(l!=m){
              cnt=0;
              ymin=min(y[l],y[m]);
              ymax=max(y[l],y[m]);
              for(o=0;o<n;o++){
                if(x[o]>=xmin&&x[o]<=xmax&&y[o]<=ymax&&y[o]>=ymin)
                  cnt++;
              }
              if(cnt>=k) s=min(s,(xmax-xmin)*(ymax-ymin));
            }
          }
        }
      }
    }
  }
  printf("%ld\n",s);
}