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<algorithm>
#include<vector>
#include<queue>
#include<string>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;
ll x[60],y[60];
pair<ll,ll> lis[60];
int main(){
	int n,k;
	cin>>n>>k;
	lol(i,n){
		cin>>x[i]>>y[i];
		lis[i]=make_pair(y[i],i);
	}
	sort(lis,lis+n);
	ll ans=(ll)mod*(ll)mod*4;
	for(int i=0;i<n;i++){
		for(int j=i+1;j<n;j++){
			ll left=lis[i].first;
			ll right=lis[j].first;
			vector<ll> v;
			for(int r=0;r<n;r++){
				if(left<=y[r]&&y[r]<=right){
					v.push_back(x[r]);
				}
			}
			sort(v.begin(),v.end());
			for(int r=k-1;r<v.size();r++){
				ans=min(ans,(v[r]-v[r-k+1])*(right-left));
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}