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 <bits/stdc++.h>

using namespace std;
using llong = long long;
using ldbl = long double;
using lpair = pair<llong, llong>;

#define ALL(x) x.begin(), x.end()

constexpr llong mod = 1e9+7;
constexpr llong inf = mod * mod;

int main() {
  llong N, K;
  vector<lpair> xy;

  cin >> N >> K;
  xy.resize(N);
  for (auto &in: xy) {
    cin >> in.first >> in.second;
  }
  sort(ALL(xy));

  llong ans = LLONG_MAX;
  for (int xl = 0; xl < N; xl++) {
    for (int xr = xl + 1; xr < N; xr++) {
      vector<int> y;
      for (int i = xl; i <= xr; i++) {
        y.push_back(xy[i].second);
      }
      sort(ALL(y));
      for (int yl = 0; yl + K - 1 < y.size(); yl++) {
        int yr = yl + K - 1;
        ans = min(ans, (xy[xr].first - xy[xl].first) * (y[yr] - y[yl]));
      }
    }
  }

  cout << ans << endl;

  return 0;
}