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>
using namespace std;
int main(){
	int n,k;
	cin >> n >> k;
  vector<pair<int,int>> p(n);
  	for(int i = 0; n > i; i++)
      	cin >> p[i].first >> p[i].second;
  
  long long int Min = -1;
  for(int i = 0; n > i; i++)
  for(int j = 0; n > j; j++)
  for(int kk = 0; n > kk; kk++)
  for(int l = 0; n > l; l++){
  	long long int top = p[i].second,btm=p[j].second,lft=p[kk].first,rgt=p[l].first;
    int count = 0;
    for(int c = 0; n > c; c++){
		if(p[c].first >= lft && p[c].first <= rgt && p[c].second >= btm && p[c].second <= top){
          count++;
        }
    }
    if(count >= k) {
     long long int area = (top - btm) * (rgt - lft);
     if(Min == -1 || Min >area) Min = area; 
    }
  }
    cout << Min << endl;
  return 0;
}