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;
typedef long long LL;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()

int main() {
  int n, k;
  cin >> n >> k;
  vector<int> x(n), y(n);
  vector<LL> nx(n), ny(n);
  rep(i, n) {
    cin >> x[i] >> y[i];
    nx[i] = x[i], ny[i] = y[i];
  }
  sort(all(nx));
  sort(all(ny));

  LL ans = (nx.back() - nx.front()) * (ny.back() - ny.front());
  rep(lx, n) for (int rx = lx + 1; rx < n; rx ++) rep(dy, n) for (int uy = dy + 1; uy < n; uy ++) {
    LL left = nx[lx], right = nx[rx];
    LL up = ny[uy], down = ny[dy];
    int num = 0;
    rep(i, n) {
      if (left <= x[i] && x[i] <= right && down <= y[i] && y[i] <= up) num ++;
    }
    if (k <= num) {
      ans = min(ans, (right - left) * (up - down));
    }
  }
  cout << ans << endl;
}