Problem: 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.

Examples

Input

4 4
1 4
3 3
6 2
8 1


Output

21


Input

4 2
0 0
1 1
2 2
3 3


Output

1


Input

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999


Output

3999999996000000001

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
long ans=8e18;
pair<long,long>x[100];
int n,k;
int main()
{
	cin>>n>>k;
	for(int i=0;i<n;i++)cin>>x[i].first>>x[i].second;
	sort(x,x+n);
	for(int i=0;i<=n-k;i++)
	{
		vector<long>P;
		for(int j=0;j<k-1;j++)
		{
			P.push_back(x[i+j].second);
		}
		for(int j=k-1;i+j<n;j++)
		{
			P.push_back(x[i+j].second);
			sort(P.begin(),P.end());
			for(int l=0;l<=P.size()-k;l++)
			{
				ans=min(ans,(max(P[l+k-1],max(x[i].second,x[i+j].second))-min(P[l],min(x[i].second,x[i+j].second)))*(x[i+j].first-x[i].first));
			}
		}
	}
	cout<<ans<<endl;
}