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 <utility>
using namespace std;
typedef long long ll;
int INF=1<<29;

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	ll N,K;
	cin>>N>>K;
	vector<pair<ll,ll>> axis(N);
	vector<ll> x(N),y(N);
	vector<ll> ans;
	for(int i=0;i<N;i++){
		cin>>x[i]>>y[i];
		axis[i].first=x[i];
		axis[i].second=y[i];
	}
	for(int i=0;i<N;i++){ //left
		for(int j=0;j<N;j++){ //right
			for(int k=0;k<N;k++){ //up
				for(int l=0;l<N;l++){ //down
					if(x[i]>=x[j] || y[l]>=y[k]) continue;
					ll c=0;
					for(int m=0;m<N;m++){
						if(x[i]<=axis[m].first && axis[m].first<=x[j] && y[l]<=axis[m].second && axis[m].second<=y[k]) c++;
					}
					if(c>=K) ans.push_back((x[j]-x[i])*(y[k]-y[l]));
				}
			}
		}
	}
	sort(ans.begin(),ans.end());
	cout<<ans[0]<<endl;
	return 0;
}