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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define INF 5000000000000000000LL
#define ll long long
typedef pair<ll,ll>P;
P t[59];

int main(){
  ll n,k;
  ll ny[59];
  cin>>n>>k;
  for(int i=0;i<n;i++){
    cin>>t[i].fi>>t[i].se;
    ny[i]=t[i].se;
  }
  sort(ny,ny+n);
  sort(t,t+n);
  ll ans=INF;
  for(int i=0;i<n;i++){
    for(int j=i;j<n;j++){
      for(int p=0;p<n;p++){
	int cnt=0;
	for(int q=p;q<n;q++){
	  if(ny[i]<=t[q].se&&t[q].se<=ny[j]){
	    cnt++;
	  }
	  if(cnt>=k){
	    ans=min(ans,(ny[j]-ny[i])*(t[q].fi-t[p].fi));
	  }
	}
      }
    }
  }
  cout<<ans<<endl;
  return(0);
}