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;

long long N, K, X[100], Y[100], minx = (1LL << 62);
vector<long long>AX, AY;

int main() {
	cin >> N >> K;
	for (int i = 1; i <= N; i++) {
		cin >> X[i] >> Y[i];
		AX.push_back(X[i]);
		AY.push_back(Y[i]);
	}
	sort(AX.begin(), AX.end());
	sort(AY.begin(), AY.end());
	for (int i = 0; i < AX.size(); i++) {
		for (int j = i; j < AX.size(); j++) {
			for (int k = 0; k < AY.size(); k++) {
				for (int l = k; l < AY.size(); l++) {
					int cnt = 0;
					for (int m = 1; m <= N; m++) {
						if (AX[i] <= X[m] && X[m] <= AX[j] && AY[k] <= Y[m] && Y[m] <= AY[l]) cnt++;
					}
					if (cnt >= K) minx = min(minx, (AX[j] - AX[i])*(AY[l] - AY[k]));
				}
			}
		}
	}
	cout << minx << endl;
	return 0;
}