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;

int main() {
	int N, K;
	long long ans = pow(2, 63) - 513;
	pair<long long, long long> P[50];
	cin >> N >> K;
	for (int i = 0; i < N; i++) {
		cin >> P[i].first >> P[i].second;
	}
	for (int i = 0; i < N; i++) {
		for (int j = i; j < N; j++) {
			for (int k = 0; k < N; k++) {
				for (int l = k; l < N; l++) {
					long long xl = min(P[i].first, P[j].first), xr = max(P[i].first, P[j].first);
					long long yu = min(P[k].second, P[l].second), yd = max(P[k].second, P[l].second);
					int count = 0;
					for (int m = 0; m < N; m++) {
						if (xl <= P[m].first && P[m].first <= xr && yu <= P[m].second && P[m].second <= yd) count++;
					}
					if (count >= K) ans = min(ans, (xr - xl) * (yd - yu));
				}
			}
		}
	}
	cout << ans;
}