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: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <string.h>

using namespace std;
typedef pair<long long int, long long int> P;

void solve()
{
	long long int N, K;
	cin >> N >> K;
	vector<long long int> x(N);
	vector<long long int> y(N);
	for (int i = 0; i < N; ++i)
	{
		cin >> x[i] >> y[i];
	}
	long long int ans = 1LL << 62;
	for (int i = 0; i < N; ++i)
	{
		for (int j = 0; j < N; ++j)
		{
			for (int k = 0; k < N; ++k)
			{
				for (int l = 0; l < N; ++l)
				{
					if (x[i] >= x[j] || y[k] >= y[l])
					{
						continue;
					}
					int cnt = 0;
					for (int m = 0; m < N; ++m)
					{
						if (x[i] <= x[m] && x[m] <= x[j] && y[k] <= y[m] && y[m] <= y[l])
						{
							++cnt;
						}
					}
					if (cnt >= K)
					{
						ans = min(ans, 1LL * (x[j] - x[i]) * (y[l] - y[k]));
					}
				}
			}
		}
	}
	cout << ans << endl;
}

int main()
{
	solve();
	return(0);
}