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 ALL(A) (A).begin(),(A).end()
#define ll long long
#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N,K;
	cin >> N >> K;
	long long ans = 1LL<<62;
	vector<ll> x(N),y(N);
	rep(i,N)cin >> x[i] >> y[i];
	rep(a,N)rep(b,N)rep(c,N)rep(d,N){
		ll left = min(x[a],x[b]);
		ll right= max(x[a],x[b]);
		ll up = max(y[c],y[d]);
		ll down = min(y[c],y[d]);
		int cnt = 0;
		rep(e,N)if(left<=x[e] && x[e] <= right && down <= y[e] && y[e] <= up)cnt++;
		if(cnt>=K)ans = min(ans,abs(left-right)*abs(up-down));
	}
	cout << ans << endl;
}