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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
void quicksort(long int a[],int s,int t){
  if(s<t){
    int i=s,j=t,p=a[(s+t)/2];
    while(1){
      while(a[i]<p)i++;
      while(p<a[j])j--;
      if(i>=j)break;
      a[i]^=a[j]^=a[i]^=a[j];
      i++;j--;
    }
    quicksort(a,s,i-1);quicksort(a,j+1,t);
  }
}
int main(){
  long int p[50][2],n,k,z=1L<<62L,i,j,l,m,o,x[50],y[50],a;
  scanf("%ld%ld",&n,&k);
  for(i=0;i<n;i++){scanf("%ld%ld",&p[i][0],&p[i][1]);x[i]=p[i][0];y[i]=p[i][1];}
  quicksort(x,0,n-1);quicksort(y,0,n-1);
  for(i=0;i<n;i++)for(j=i;j<n;j++)for(l=0;l<n;l++)for(m=l;m<n;m++){
    a=0;for(o=0;o<n;o++)if(p[o][0]>=x[i]&&p[o][0]<=x[j]&&p[o][1]>=y[l]&&p[o][1]<=y[m])a++;
    if(a>=k&&(long long int)(x[j]-x[i])*(y[m]-y[l])<z)z=(long long int)(x[j]-x[i])*(y[m]-y[l]);
  }
  printf("%ld",z);
}