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>
using namespace std;

typedef long long ll;
int N,K;
vector<pair<ll, ll>>v;

int main(){
	cin>>N>>K;
	for(int i=0;i<N;i++){
		ll x,y;
		cin>>x>>y;
		v.push_back(make_pair(x,y));
	}
	sort(v.begin(),v.end());
	ll ans=4*1000000000000000000;
	for(int i=0;i<v.size()-1;i++){
		for(int j=i+1;j<v.size();j++){
			int num=j-i+1;
			if(num<K)continue;
			vector<ll>y;
			for(int k=i;k<=j;k++){
				y.push_back(v[k].second);
			}
			sort(y.begin(),y.end());
			ll ydis=10000000000;
			for(int k=0;k<=y.size()-K;k++){
				ydis=min(ydis,y[k+K-1]-y[k]);
			}
			ll xdis=v[j].first-v[i].first;
			ans=min(ans,xdis*ydis);
		}
	}
	cout<<ans<<endl;
	return 0;
}