Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4 4
1 4
3 3
6 2
8 1

Sample Output 1

21

One rectangle that satisfies the condition with the minimum possible area has the following vertices: (1,1), (8,1), (1,4) and (8,4).

Its area is (8-1) × (4-1) = 21.

Sample Input 2

4 2
0 0
1 1
2 2
3 3

Sample Output 2

1

Sample Input 3

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999

Sample Output 3

3999999996000000001

Watch out for integer overflows.

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

def solve(xt_o,yt_o,pt,n,k)
  xt = xt_o.sort
  yt = yt_o.sort
  target = []
  ans = Float::INFINITY

  for x1 in 0..(n-1) do
    for x2 in (x1+1)..(n-1) do
      for y1 in 0..(n-1) do
        for y2 in (y1+1)..(n-1) do
          target << [xt[x1],yt[y1],xt[x2],yt[y2]]
        end
      end
    end
  end
#cumulative sum
  arr = Hash.new(0)
  for x in 0..(n-1) do
    for y in 0..(n-1) do
      cnt = 0
      for i in 0..(n-1) do
        judge = ((xt[0] <= xt_o[i]) && (xt_o[i] <= xt[x])) &&
                ((yt[0] <= yt_o[i]) && (yt_o[i] <= yt[y]))
        cnt = cnt + 1 if judge
      end
    arr[ [xt[x],yt[y]] ] = cnt
    end
  end
  target.each{|t|
  corner = pt[ [t[0],t[1]] ] + pt[ [t[0],t[3]] ] + pt[ [t[2],t[1]] ]
  cnt = arr[ [t[2],t[3]] ] - arr[ [t[0],t[3]] ] - arr[ [t[2],t[1]] ] + arr[ [t[0],t[1]] ] + corner
  if k <= cnt then
    area = (t[2] - t[0]).abs * (t[3] - t[1]).abs
    ans = [ans,area].min
  end
}
ans
end

def main()
  n,k = gets.chomp.split(' ').map(&:to_i)
  xtable = []
  ytable = []
  dic = Hash.new(0)
  for num in 0..n-1 do
    x,y = gets.chomp.split(' ').map(&:to_i)
    xtable << x
    ytable << y
    dic[[x,y]] = 1
  end
  puts solve(xtable,ytable,dic,n,k)
end

main()