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<string>
#include<cmath>
#include<vector>
#include<queue>
#include<functional>
#include<map>
using namespace std;
#pragma warning(disable: 4996)
vector<long long>X, Y; long long N, K, x[1000], y[1000], minx = (1LL << 62);
int main() {
	cin >> N >> K;
	for (int i = 0; i < N; i++) { cin >> x[i] >> y[i]; X.push_back(x[i]); Y.push_back(y[i]); }
	sort(X.begin(), X.end()); sort(Y.begin(), Y.end());
	for (int i = 0; i < X.size(); i++) {
		for (int j = i + 1; j < X.size(); j++) {
			vector<long long>Z;
			for (int k = 0; k < N; k++) {
				if (X[i] <= x[k] && x[k] <= X[j])Z.push_back(y[k]);
			}
			sort(Z.begin(), Z.end());
			for (int k = 0; k <= (int)Z.size() - K; k++) {
				minx = min(minx, (X[j] - X[i])*(Z[k + K - 1] - Z[k]));
			}
		}
	}
	cout << minx << endl;
	return 0;
}