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: Accepted)

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

    class Program
    {
        static void Main(string[] args)
        {
            string line1 = Console.ReadLine();
            string line2 = Console.ReadLine();


            List<long> NK = new List<long>(line1.Split(' ').Select((arg) => long.Parse(arg)));
            List<long> RowAs = new List<long>(line2.Split(' ').Select((arg) => long.Parse(arg)));

            long inCount = 0;
            long outCount = 0;

            for (int i = 0; i < RowAs.Count; i++)
            {
                if (i + 1 < RowAs.Count)
                {
                    for (int k = i+1; k < RowAs.Count; k++)
                    {
                        if (RowAs[i] > RowAs[k])
                        {
                            inCount++;
                        }
                    }
                }
            }



            for (int i = 0; i < RowAs.Count; i++)
            {
                    for (int k = 0; k < RowAs.Count; k++)
                    {
                        if (RowAs[i] > RowAs[k])
                        {
                            outCount++;
                        }
                    }
            }

            long mod = (long)(Math.Pow(10, 9) + 7);

            inCount = (NK[1] * inCount) % mod;
            outCount = ((((NK[1] * (NK[1] - 1)) / 2) % mod) * outCount) % mod;



            Console.WriteLine((inCount+outCount)%mod);

        }
    }