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: Accepted)

#include <iostream>
#include <vector>

using namespace std;

using int64 = long long;

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

    vector<int64> x(n);
    vector<int64> y(n);
    for (int i=0; i<n; i++) cin >> x[i] >> y[i];

    vector<int64> xv = x;
    vector<int64> yv = y;

    sort(xv.begin(), xv.end());
    sort(yv.begin(), yv.end());

    int64 ans = LONG_MAX;
    for (int xs=0; xs<n; xs++) {
        for (int xe=xs+1; xe<n; xe++) {
            for (int ys=0; ys<n; ys++) {
                for (int ye=ys+1; ye<n; ye++) {

                    int cnt = 0;
                    for (int i=0; i<n; i++) {
                        if (xv[xs] <= x[i] && x[i] <= xv[xe] && yv[ys] <= y[i] && y[i] <= yv[ye])
                            cnt++;
                    }
                    if (cnt >= k) {
                        ans = min(ans, (xv[xe] - xv[xs]) * (yv[ye] - yv[ys]));
                    }
              }
            }
        }
    }

    cout << ans << endl;
}