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:
//in the name of god
#include <bits/stdc++.h>
using namespace std;
#define MAX_N (int)1e5+20
#define INF 1e18+20
#define pb push_back
#define F first
#define S second
typedef long long LL;
typedef pair<LL,LL> pii;
const int delta=(int)1e9+7;
LL n,k,ans=-1,a,b;pii x[70]; 
int main(){
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin>>n>>k;
	for(int i=0;i<n;++i) cin>>x[i].F>>x[i].S;
	sort(x,x+n);
	for(int i=0;i<n;++i){
		for(int j=i;j<n;++j){
			a=x[j].F-x[i].F;
			vector<int>y; y.clear();
			for(int l=i;l<=j;++l){
				y.pb(x[l].S);
			}
			if(y.size()<k) continue;
			sort(y.begin(),y.end());
			b=INF;
			for(int l=k-1;l<y.size();++l){
				b=min(b,(LL)(y[l]-y[l-k+1]));
			}
			//cout<<a<<" "<<b<<" "<<a*b<<endl;
			if(ans==-1) ans=a*b;
			ans=min(ans,a*b);
		}
	}
	cout<<ans;
	return cout<<endl,0;
}