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 <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <math.h>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<ll,ll>;

int main() {
  int n,k;
  cin >> n >> k;
  vector<P> v(n);
  rep(i,n) cin >> v[i].first >> v[i].second;
  sort(v.begin(),v.end());

  ll ans = 5L*1e18L;
  rep(l,n-k+1){
    vector<ll> y;
    rep(i,k-1) y.push_back(v[l+i].second);
    for(int r=l+k-1;r<n;++r){
      y.push_back(v[r].second);
      sort(y.begin(),y.end());
      ll dx = v[r].first - v[l].first;
      ll dy = 1e18L;
      rep(i,y.size()-k+1) dy = min(dy,y[i+k-1]-y[i]); 
      ans = min(ans,dx*dy);
    }
  }
  cout << ans << endl;
}