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<climits>
#include<vector>
#include<queue>
#include<functional>
#include<numeric>

using namespace std;
int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n, k, x, y;
	cin >> n >> k;
	vector<pair<long long, long long>> xy(n);
	for (auto& e : xy) {
		cin >> x >> y;
		e=make_pair(x, y);
	}
	sort(xy.begin(), xy.end());
	long long ans = LLONG_MAX;

	for (int i = 0; i < n - k + 1; i++) {
		for (int j = i + k - 1; j < n; j++) {
			vector<long long> vec;
			for (int ii = i; ii <= j; ii++) {
				vec.push_back(xy[ii].second);
			}
			sort(vec.begin(), vec.end());
			for (int ii = 0; ii <= j - i - k + 1; ii++) {
				ans = min(ans, (vec[ii + k-1] - vec[ii])*(xy[j].first-xy[i].first));
			}
		}
	}

	cout << ans << endl;

	return 0;
}