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 ConsoleApp1
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] inputs = Console.ReadLine().Split(' ');
            int N = int.Parse(inputs[0]);
            long K = long.Parse(inputs[1]);
            string[] alist = Console.ReadLine().Split(' ');
            int[] a = new int[N];

            for(int i = 0; i<N; i++)
            {
                a[i] = int.Parse(alist[i]);
            }

            long counter = 0;

            long small = 0;
            long large = 0;

            for(int i=0; i<N -1; i++)
            {
                for(int j=i+1; j<N; j++)
                {
                    if(a[i] == a[j]) { continue; }
                    if(a[i] > a[j])
                    {
                        large++;
                    }
                    else
                    {
                        small++;
                    }
                }
            }

            //Console.WriteLine(K);
            //Console.WriteLine(int.MaxValue());
            //Console.WriteLine("Large {0} Small {1}", large, small);
            counter += large * K;
            counter += (large + small) * (K * (K - 1)) / 2;
            long tmp = 1000000007;

            Console.WriteLine(counter%tmp);
            //Console.ReadKey();

        }

    }


}