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>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;


int main(void) {
	int n, k, i, j, l, x, y;
	ll ans = 1e20, s, h;
	vector<P> p;

	cin >> n >> k;
	for (i = 0; i < n; i++) {
		cin >> x >> y;
		p.push_back(P(x, y));
	}

	sort(p.begin(), p.end());

	for (i = 0; i < n; i++) for (j = i + k - 1; j < n; j++) {
		vector<ll> yp;
		for (l = i; l <= j; l++) {
			yp.push_back(p[l].second);
		}
		sort(yp.begin(), yp.end());
		h = 2e9;
		for (l = 0; l + k - 1 < j - i + 1; l++) {
			h = min(h, yp[l + k - 1] - yp[l]);
		}
		s = ((ll)p[j].first - p[i].first) * h;
		ans = min(ans, s);
	}

	cout << ans << endl;

	return 0;
}