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 mod 1000000007
#define ll long long int
#define ar array
#define pb push_back
using namespace std;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	int n,k;
	cin>>n>>k;
	vector<int> x(n), y(n);
	for(int i = 0; i<n; i++)
	{
		cin>>x[i];
		cin>>y[i];
	}
	vector<int> x3(x),y3(y);
	sort(x.begin(),x.end());
	sort(y.begin(),y.end());
	ll l1,r1,l2,r2,ans = (ll)1e20;
	for(int lx = 0; lx<n; lx++)
		for(int rx = lx+1;rx<n; rx++)
			for(int ly = 0; ly<n; ly++)
				for(int ry = ly+1; ry<n; ry++)
				{
					l1 = x[lx];
					l2 = x[rx];
					r1 = y[ly];
					r2 = y[ry];
					int num = 0;
					for(int i = 0; i<n; i++)
					{
						if(l1<=x3[i] and l2>=x3[i] and r1<=y3[i] and r2>=y3[i])
							num++;
					}
					if(num>=k) ans = min(ans,(l2-l1)*(r2-r1));
				}
	cout<<ans<<endl;
	return 0;
}