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;
typedef pair<ll, ll> P;

int main() {
	int N, K; cin >> N >> K;
	vector<ll> x(N), y(N);
	vector<P> xy(N);
	for (int i = 0; i < N; i++) {
		cin >> x[i] >> y[i];
		xy[i] = P(x[i], y[i]);
	}
	sort(x.begin(), x.end());
	sort(y.begin(), y.end());

	ll ans = (x[N - 1] - x[0])*(y[N - 1] - y[0]);
	for (int x0 = 0; x0 < N - 1; x0++) {
		for (int x1 = x0 + 1; x1 < N; x1++) {
			for (int y0 = 0; y0 < N - 1; y0++) {
				for (int y1 = y0 + 1; y1 < N; y1++) {
					int cnt = 0;
					for (int i = 0; i < N; i++) {
						if (x[x0] <= xy[i].first && xy[i].first <= x[x1] && y[y0] <= xy[i].second && xy[i].second <= y[y1]) {
							cnt++;
						}
					}

					if (K <= cnt) {
						ans = min(ans, (x[x1] - x[x0])*(y[y1] - y[y0]));
					}
				}
			}
		}
	}

	cout << ans << endl;
}