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;

int main()
{
    int n, k; cin>>n>>k;
	vector<int> x(n), y(n);
	for(int i=0; i<n; ++i) cin>>x[i]>>y[i];

    long res=LONG_MAX;
#define rep(i, a, b) for(int i=a; i<b; ++i)
	rep(a, 0, n) rep(b, a, n) rep(c, b, n) rep(d, c, n) {
		auto ax=minmax({x[a], x[b], x[c], x[d]});
		auto ay=minmax({y[a], y[b], y[c], y[d]});
		long minx=ax.first, maxx=ax.second;
		long miny=ay.first, maxy=ay.second;
		int cnt=0;
		rep(e, 0, n)
			cnt+=(minx<=x[e]&&x[e]<=maxx)&&(miny<=y[e]&&y[e]<=maxy);
		if (cnt>=k)
			res=min(res, (maxx-minx)*(maxy-miny));
	}
    cout<<res<<endl;
}