Problem: 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.

Examples

Input

4 4
1 4
3 3
6 2
8 1


Output

21


Input

4 2
0 0
1 1
2 2
3 3


Output

1


Input

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999


Output

3999999996000000001

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstdint>
#include <vector>
#include <algorithm>
#include <utility>
#include <tuple>

int main() {
  size_t N, K;
  scanf("%zu %zu", &N, &K);

  std::vector<std::pair<intmax_t, intmax_t>> xy(N);
  std::vector<intmax_t> xs(N), ys(N);
  for (size_t i = 0; i < N; ++i) {
    scanf("%jd %jd", &xy[i].first, &xy[i].second);
    std::tie(xs[i], ys[i]) = xy[i];
  }

  intmax_t res = 9e18;
  for (auto x0: xs)
    for (auto x1: xs)
      for (auto y0: ys)
        for (auto y1: ys) {
          size_t cur = 0;
          for (const auto& xyi: xy) {
            intmax_t x, y;
            std::tie(x, y) = xyi;
            if (x0 <= x && x <= x1 && y0 <= y && y <= y1) ++cur;
          }

          if (cur < K) continue;
          res = std::min(res, (y1-y0)*(x1-x0));
        }

  printf("%jd\n", res);
}