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 <bits/stdc++.h>
#define pb push_back
#define int long long
#define mp make_pair
#define inf 1000000007
#define LINF 6000000000000000007LL
#define ll long long
using namespace std;
signed main(){
	int n,k;
	cin>>n>>k;
	vector< pair<int,int> > a;
	for(int i=0;i<n;i++){
		int x,y;
		cin>>x>>y;
		a.pb(mp(x,y));
	}
	sort( a.begin(), a.end() );
	ll ans = LINF;
	for(int i=0;i<=n-k;i++){
		for(int j=i+k-1;j<n;j++){
			int lx = a[j].first - a[i].first;
			vector<int> ys;
			for(int p=i;p<=j;p++){
				ys.pb( a[p].second );
			}
			sort( ys.begin(),ys.end() );
			int ly = LINF;
			for(int p=0;p<=ys.size()-k;p++){
				ly = min( ly, ys[p+k-1] - ys[p]);
			}
			ll tmp = lx * ly;
			if( ans > tmp ) ans = tmp;
		}
	}
	cout<<ans<<endl;
	return 0;
}