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;
 
struct Point{int x, y;} p[55];
bool lessx(Point &a, Point &b){return a.x < b.x;}

int main(){
  int n, k, v[55], i, j, l, m, cnt;
  long long ans = 4e18;
  scanf("%d%d", &n, &k);
  for(i = 0; i < n; i++){
    scanf("%d%d", &p[i].x, &p[i].y);
    v[i] = p[i].y;
  }
  sort(p, p+n, lessx);
  sort(v, v+n);
  for(i = 0; i <= n-k; i++){
    for(j = i+k-1; j < n; j++){
      for(l = 0; l <= n-k; l++){
        cnt = 0;
        for(m = l; m < n && cnt < k; m++) if(v[i] <= p[m].y && p[m].y <= v[j]) cnt++;
        if(cnt < k) break;
        ans = min(ans, ((long long) v[j] - v[i])*(p[m-1].x - p[l].x));
      }
    }
  }
  printf("%lld\n", ans);
}