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 <cstdio>
#include <climits>
#include <algorithm>

using namespace std;

int const N_MAX = 50;

int N, K;
long long xs[N_MAX + 1], ys[N_MAX + 1];

int main() {
  scanf("%d %d", &N, &K);
  for (int i = 0; i < N; i++) {
    scanf("%lld %lld", &xs[i], &ys[i]);
  }

  long long area = LLONG_MAX;
  for (int xi = 0; xi < N; xi++){
    for (int xj = 0; xj < N; xj++) {
      if (xi == xj) continue;

      long long xl = min(xs[xi], xs[xj]);
      long long xr = max(xs[xi], xs[xj]);

      for (int yi = 0; yi < N; yi++) {
        for (int yj = 0; yj < N; yj++) {
          if (yi == yj) continue;

          long long yd = min(ys[yi], ys[yj]);
          long long yu = max(ys[yi], ys[yj]);

          int count = 0;
          for (int i = 0; i < N; i++) {
            int x = xs[i];
            int y = ys[i];

            if ((x >= xl && x <= xr) && (y >= yd && y <= yu)) count++;
          }
          if (count == K) {
            long long new_area = (xr - xl) * (yu - yd);
            if (new_area < area) area = new_area;
          }
        }
      }
    }
  }
  printf("%lld\n", area);
}