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 <algorithm>
#include <cstring>
#include <deque>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <vector>
using namespace std;
using ll = long long;

int main() {
  int N, K;
  cin >> N >> K;
  vector<ll> x(N), y(N);
  for (int i = 0; i < N; i++) {
    cin >> x[i] >> y[i];
  }
  ll res = 1ll << 62;
  for (int i = 0; i < N; i++) {
    for (int j = i + 1; j < N; j++) {
      ll xs = x[i], xt = x[j];
      if (xs > xt) swap(xs, xt);
      vector<ll> ny;
      for (int k = 0; k < N; k++) {
        if (xs <= x[k] && x[k] <= xt) {
          ny.push_back(y[k]);
        }
      }
      sort(ny.begin(), ny.end());
      for (int k = 0; k + K - 1 < ny.size(); k++) {
        res = min(res, (xt - xs) * (ny[k + K - 1] - ny[k]));
      }
    }
  }
  cout << res << endl;
  return 0;
}