Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4 4
1 4
3 3
6 2
8 1

Sample Output 1

21

One rectangle that satisfies the condition with the minimum possible area has the following vertices: (1,1), (8,1), (1,4) and (8,4).

Its area is (8-1) × (4-1) = 21.

Sample Input 2

4 2
0 0
1 1
2 2
3 3

Sample Output 2

1

Sample Input 3

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999

Sample Output 3

3999999996000000001

Watch out for integer overflows.

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

int main() {
	int n, need;
	cin >> n >> need;
	int64_t x[n], y[n];
	for (int i = 0; i < n; i++) {
		cin >> x[i] >> y[i];
	}
	int64_t ans = numeric_limits<int64_t>::max();
	if (need == 2) {
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				if (i == j) continue;
				ans = min(ans, (max(x[i], x[j]) - min(x[i], x[j])) * (max(y[i], y[j]) - min(y[i], y[j])));
			}
		}
	} else {
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				for (int k = 0; k < n; k++) {
					if (i == j || j == k || i == k) continue;
					int count = 0;
					int64_t left = min({x[i], x[j], x[k]});
					int64_t right = max({x[i], x[j], x[k]});
					int64_t down = min({y[i], y[j], y[k]});
					int64_t up = max({y[i], y[j], y[k]});
					for (int l = 0; l < n; l++) {
						if (left <= x[l] && x[l] <= right && down <= y[l] && y[l] <= up) {
							count++;
						}
					}
					if (need <= count) {
						ans = min(ans, (right - left) * (up - down));
					}
				}
			}
		}
	}
	assert(ans != numeric_limits<int64_t>::max());
	cout << ans << endl;
	return 0;
}