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;
using System.Collections.Generic;
using System.Linq;

namespace AtCoder
{
    class Program
    {
        static void Main(string[] args)
        {
            var array = Console.ReadLine().Split(' ').Select(_ => int.Parse(_));
            var N = array.First();
            var K = array.Last();

            var A = Console.ReadLine().Split(' ').Select(_ => int.Parse(_)).ToArray();

            var z = ((decimal)Math.Pow(10, 9)) + 7;
            decimal tentouSuuA = 0;
            for (int i = 0; i < N - 1; i++)
            {
                for (int j = i + 1; j < N; j++)
                {
                    var Ai = A[i];
                    var Aj = A[j];
                    if (A[i] > A[j])
                    {
                        if (K == 1)
                        {
                            tentouSuuA += 1;
                        }
                        else
                        {
                            tentouSuuA += ((K % z) * K - 1) % z;
                        }
                    }
                }
            }

            if (K > 1)
            {
                for (int i = 0; i < N - 1; i++)
                {
                    if (A[N - 1] > A[i])
                    {
                        tentouSuuA += (((K-1) % z) * (K - 1)) % z;
                    }
                }
            }

            Console.WriteLine(tentouSuuA % z);
        }
    }
}