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;
int main() {
  long long N, K, answer = LLONG_MAX;
  cin >> N >> K;
  pair<int,int> xy[N];
  for (int i = 0; i < N; i++) {
    cin >> xy[i].first >> xy[i].second;
  }
  sort(xy,xy+N);
  for (int i = 0; i < N; i++) {
    for (int j = i+K-1; j < N; j++) {
      int z[j-i+1];
      for (int k = i; k <= j; k++) {
        z[k-i] = xy[k].second;
      }
      sort(z,z+j-i+1);
      for (int k = 0; k+K-1 <= j-i; k++) {
        answer = min(answer,1ll*(xy[j].first-xy[i].first)*(z[k+K-1]-z[k]));
      }
    }
  }
  cout << answer;
}