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<algorithm>
#include<vector>
#include<climits>
using namespace std;
int N_MAX = 50;
typedef pair<long long,long long> P;
int main()
{
  int n,k,i,j,l;
  long long x,y,res=LLONG_MAX;
  P p[N_MAX];
  vector<long long> q;
  cin >> n >> k;
  for(i=0;i<n;i++)
    cin >> p[i].first >> p[i].second;
  sort(p,p+n);
  for(i=0;i<n-k+1;i++){
    q.clear();
    x=p[i+k-1].first-p[i].first;
    for(j=i;j<i+k;j++)
      q.push_back(p[j].second);
    sort(q.begin(),q.end());
    y=q[k-1]-q[0];
    res=min(res,x*y);
    for(j=i+k;j<n;j++){
      q.push_back(p[j].second);
      sort(q.begin(),q.end());
      x=p[j].first-p[i].first;
      y=LLONG_MAX;
      for(l=0;l<=j-i-k+1;l++)
        y=min(y,q[l+k-1]-q[l]);
      res=min(res,x*y);
    }
  }
  cout << res << endl;
  return 0;
}