Problem:
Score : 300 points

Problem Statement

We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}.

Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2.

Find the inversion number of B, modulo 10^9 + 7.

Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.

Constraints

All values in input are integers.

1 \leq N \leq 2000

1 \leq K \leq 10^9

1 \leq A_i \leq 2000

Input

Input is given from Standard Input in the following format:

N K
A_0 A_1 ... A_{N - 1}

Output

Print the inversion number of B, modulo 10^9 + 7.

Sample Input 1

2 2
2 1

Sample Output 1

3

In this case, B~=~2,~1,~2,~1. We have:

B_0 > B_1

B_0 > B_3

B_2 > B_3

Thus, the inversion number of B is 3.

Sample Input 2

3 5
1 1 1

Sample Output 2

0

A may contain multiple occurrences of the same number.

Sample Input 3

10 998244353
10 9 8 7 5 6 3 4 2 1

Sample Output 3

185297239

Be sure to print the output modulo 10^9 + 7.

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

using System;

namespace pro
{
    class MainClass
    {
        public static void Main()
        {
            string[] str1 = Console.ReadLine().Split(' ');
            int n = int.Parse(str1[0]);
            long k = int.Parse(str1[1]);
            int[] a = new int[n];
            string[] str2 = Console.ReadLine().Split(' ');
            for (int i = 0; i < n; i++)
            {
                a[i] = int.Parse(str2[i]);
            }


            int count = 0;
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    if (a[i] > a[j])
                    {
                        count++;
                    }
                }
            }
            int count1 = 0;
            for (int i = 0; i < n; i++)
            {
                for (int j = i + 1; j < n; j++)
                {
                    if (a[i] > a[j])
                    {
                        count1++;
                    }
                }
            }
                long sum = 0;

            sum = (long)(count * ((k * k - k) % (long)(Math.Pow(10, 9) + 7) / 2.0));
                sum =( sum % (long)(Math.Pow(10, 9) + 7) + count1 * k % (long)(Math.Pow(10, 9) + 7) )% (long)(Math.Pow(10, 9) + 7);
                Console.WriteLine(sum);
            

        }
    }

}