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;
long long INF = 4000000000000000000;
int main(){
  int N, K;
  cin >> N >> K;
  vector<int> x(N), y(N);
  for (int i = 0; i < N; i++){
    cin >> x[i] >> y[i];
  }
  long long ans = INF;
  for (int i = 0; i < N; i++){
    for (int j = 0; j < N; j++){
      vector<int> y2;
      for (int k = 0; k < N; k++){
        if (x[i] <= x[k] && x[k] <= x[j]){
          y2.push_back(y[k]);
        }
      }
      sort(y2.begin(), y2.end());
      if (y2.size() >= K){
        long long W = abs(x[j] - x[i]);
        int cnt = y2.size();
        for (int i = 0; i < cnt - K + 1; i++){
          long long H = y2[i + K - 1] - y2[i];
          ans = min(ans, H * W);
        }
      }
    }
  }
  cout << ans << endl;
}