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 <bits/stdc++.h>

using namespace std;

using ll = long long;

int main() {
  int n, k;
  cin >> n >> k;
  vector<ll> x(n), y(n);
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i];
  }
  vector<ll> sx(x), sy(y);
  sort(x.begin(), x.end());
  sort(y.begin(), y.end());
  ll ans = (x[n - 1] - x[0]) * (y[n - 1] - y[0]);
  for (int x_i = 0; x_i < n; ++x_i) {
    for (int x_j = x_i + 1; x_j < n; ++x_j) {
      for (int y_i = 0; y_i < n; ++y_i) {
        for (int y_j = y_i + 1; y_j < n; ++y_j) {
          int count = 0;
          for (int i = 0; i < n; ++i) {
            if (x[x_i] <= sx[i] && x[x_j] >= sx[i] && y[y_i] <= sy[i] &&
                y[y_j] >= sy[i]) {
              ++count;
            }
          }
          if (count >= k) {
            ll ret = (x[x_j] - x[x_i]) * (y[y_j] - y[y_i]);
            if (ret)
              ans = min(ans, ret);
          }
        }
      }
    }
  }
  cout << ans << endl;
}