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;
#define all(V) V.begin(),V.end()
#define ll long long
#define MOD 1000000007

int main() {
	int N, K, a, b, ts;
	cin >> N >> K;
	vector<pair<ll, ll>> Xf(N);
	for (int i = 0;i < N;i++) {
		cin >> a >> b;
		Xf[i] = { a,b };
	}
	sort(all(Xf));
	vector<ll> T, S;
	ll ymi, xmi, ans = LLONG_MAX;
	for (int i = 0;i < N;i++) {
		for (int j = i + 1;j < N;j++) {
			T = S;
			xmi = Xf[j].first - Xf[i].first;
			ymi = LLONG_MAX;
			for (int k = i;k <= j;k++) {
				T.push_back(Xf[k].second);
			}
			ts = T.size();
			if (ts < K)continue;
			sort(all(T));
			for (int k = 0;k + K <= ts;k++) {
				ymi = min(T[k + K - 1] - T[k], ymi);
			}
			ans = min(ans, xmi*ymi);
		}
	}
	cout << ans << endl;
}