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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;//リストの使用
using System.Collections.Generic;
using System.Text;//テキストの高速出力に必要
class Program
{
	static void Main()
	{
		string[] input = Console.ReadLine().Split(' ');//Splitで区切り文字を指定して複数個受け取る。
		long n = long.Parse(input[0]);
		long k = long.Parse(input[1]);
    long answer = long.MaxValue;

    long[,] vertexes = new long[n,2];//ここに受け取る
    for(long i = 0; i < n; i++)
    {
      long[] nums = Array.ConvertAll(Console.ReadLine().Split(' '),long.Parse);//１行ごとに受け取る
      vertexes[i,0] = nums[0];
      vertexes[i,1] = nums[1];
    }

    for(int xa = 0; xa < n; xa++)
    {
      for(int xb = 0; xb < n; xb++)
      {
        for(int ya = 0; ya < n; ya++)
        {
          for(int yb = 0; yb < n; yb++)
          {
            int memo = 0;
            for(int i = 0; i < n; i++)
            {
              if(vertexes[i,0]<=xa && xb<=vertexes[i,0] &&
                vertexes[i,1]<=ya && yb<=vertexes[i,1]) memo++;
            }
            if(memo >= k) answer = Math.Min(answer, (xb-xa)*(yb-ya));
          }
        }
      }
    }

		Console.WriteLine(answer);//WriteLineをWriteとすると、改行なしで出力。
	}
}