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;
using System.Text;
using System.Threading.Tasks;

namespace AtCoderPre
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] s = Console.ReadLine().Split(' ');
            int N = int.Parse(s[0]);
            int K = int.Parse(s[1]);
             
            var A = Array.ConvertAll(Console.ReadLine().Split(' '), long.Parse);

            var invert = new long[N];
            for (int i = 0; i < N; i++)
                for (int j = i + 1; j < N; j++)
                    if (A[i] > A[j])
                        invert[i]++;

            var lessCount = new long[N];
            for (int i = 0; i < N; i++)
                for (int j = 0; j < N; j++)
                    if (i != j && A[i] > A[j])
                        lessCount[i]++;

            long result = 0;
            for (int i=0; i < N; i++)
            {
                result = (result + invert[i] * K);
                result = (result + lessCount[i] * (K * (K - 1)) / 2);
            }

            Console.WriteLine(result % 1000000007);
        }

    }
}