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<iostream>
#include<vector>
#include<utility>
#include<algorithm>
using namespace std;
int main(){
  long long n,k,num,min=3999999996990000001,y_length,y_min;
  cin >> n >> k;
  vector<pair<int,int>> p(n);
  for(int i=0;i<n;i++){
    cin >> p[i].first >> p[i].second;
  }
  sort(p.begin(),p.end());
  for(int r=0;r<n;r++){
    for(int l=r+k-1;l<n;l++){
      vector<int> tmp;
      for(int i=r;i<=l;i++){
        tmp.push_back(p[i].second);
      }
      sort(tmp.begin(),tmp.end());
      y_min = tmp[k-1] - tmp[0];
      for(int yr=0;yr<tmp.size();yr++){
        for(int yl=yr+k-1;yl<tmp.size();yl++){
          y_length = tmp[yl]-tmp[yr];
          y_min = (y_min<y_length)?y_min:y_length;
        }
      }
      num = y_min*(p[l].first-p[r].first);
      min = (num<min)?num:min;
    }
  }
  cout << min << endl;


}