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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()

int main() {
  int n, k;
  cin >> n >> k;
  vector<int> x(n), y(n);
  vector<int> nx(n), ny(n);
  rep(i, n) {
    cin >> x[i] >> y[i];
    nx[i] = x[i];
    ny[i] = y[i];
  }

  sort(all(nx));
  sort(all(ny));

  LL ans = ((LL)nx.back() - nx.front()) * (ny.back() - ny.front());
  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 ++) {
          int num = 0;
          LL left = nx[xi], right = nx[xj];
          LL down = ny[yi], up = ny[yj];
          rep(i, n) {
            if (nx[xi] <= nx[i] && nx[i] <= nx[xj] && ny[yi] <= ny[i] && ny[i] <= ny[yj]) num ++;
          }
          if (k <= num) {
            ans = min(ans, ((LL)right - left) * (up - down));
          }
        }
      }
    }
  }

  cout << ans << endl;
}