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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;
using ll = long long;

int main()
{
    int n, k;
    cin >> n >> k;

    vector<ll> x(n), y(n);
    vector<ll> xaray, yaray;

    for (int i = 0; i < n;i++)
    {
        cin >> x[i] >> y[i];

        xaray.push_back(x[i]);
        yaray.push_back(y[i]);
    }

    sort(xaray.begin(), xaray.end());
    sort(yaray.begin(), yaray.end());

    ll ans = (ll)(xaray[n - 1] - xaray[0]) * (yaray[n - 1] - yaray[0]);

    for (int xi = 0; xi < n;xi++)
    {
        for (int xj = xi + 1; xj < n;xj++)
        {
            for (int yi = 0; yi < n;yi++)
            {
                for (int yj = yi + 1; yj < n;yj++)
                {
                    ll l = xaray[xi];
                    ll r = xaray[xj];
                    ll d = xaray[yi];
                    ll u = xaray[yj];

                    // check the number of points in the rectangle.
                    int cnt = 0;
                    for (int i = 0; i < n; i++)
                    {
                        if(l <= x[i] && x[i] <= r && d <= y[i] && y[i] <= u)
                        {
                            cnt++;
                        }
                    }

                    if(cnt >= k)
                    {
                        ans = min(ans, (ll)(r - l) * (u - d));
                    }
                }
            }
        }
    }
    cout << ans << endl;

    return 0;
}