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 lint = long long;
using VI = vector<int>;
using VVI = vector<VI>;

int main() {
  cin.tie(0); ios::sync_with_stdio(false);
  int n, k; cin >> n >> k;
  VI x(n), y(n); for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
  lint res = 9e18;
  for (int xl = 0; xl < n; xl++) {
    for (int xr = 0; xr < n; xr++) {
      for (int yl = 0; yl < n; yl++) {
        for (int yr = 0; yr < n; yr++) {
          int t = 0;
          for (int i = 0; i < n; i++) {
            t += x[xl] <= x[i] and x[i] <= x[xr] and y[yl] <= y[i] and y[i] <= y[yr];
          }
          if (t >= k) {
            res = min(res, (lint) (x[xr] - x[xl]) * (y[yr] - y[yl]));
          }
        }
      }
    }
  }
  cout << res << '\n';
  return 0;
}