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:
/*input
4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999
*/
#include <bits/stdc++.h>
using namespace std;
int n,k;
pair<long long,long long> a[1000005];
long long check(int x, int y){
	long long l=a[y].first-a[x].first,w=1e18;
	vector<long long> v;
	for(int i=x;i<=y;++i)
		v.push_back(a[i].second);
	sort(v.begin(),v.end());
	for(int i=k-1;i<v.size();++i)
		w=min(w,v[i]-v[i-k+1]);
	return l*w;
}
int main(){
	#ifdef ONLINE_JUDGE
	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	#endif
	long long ans=5e18,sum=0,p,x,y;
	cin>>n>>k;
	for(int i=0;i<n;++i){
		cin>>a[i].first>>a[i].second;
	}
	sort(a,a+n);
	for(int i=0;i<n;++i)
		for(int j=i+k-1;j<n;++j)
			ans=min(ans,check(i,j));
	cout<<ans;
}