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 static System.Console;
using System.Collections.Generic;
using System.Linq;

namespace Atcoder
{
    class Program
    {
        const double MOD = 1000000007;
        static void Main(string[] args)
        {
            var nk = ReadLine().Split(' ').Select(int.Parse).ToList();
            var a = ReadLine().Split(' ').Select(int.Parse).ToList();

            int cnt = 0; //a一個分の要素に対して
            for (int i = 0; i < a.Count(); i++)
            {
                for (int j = i + 1; j < a.Count(); j++)
                {
                    if (a[i] > a[j])
                    {
                        cnt++;
                        break;
                    }
                }
            }

            int gm = 0;
            for (int i = 1; i <= cnt; i++)
            {
                gm += i;
            }

            a.Sort();
            a.Reverse();

            int cnt_loop = 0; //くり返し用
            for (int i = 0; i < a.Count(); i++)
            {
                int tmp = 0;
                for (int j = i + 1; j < a.Count(); j++)
                {
                    if (a[i] == a[j])
                    {
                        tmp++;
                    }
                    else
                    {
                        break;
                    }
                }
                cnt_loop += a.Count() - i - 1 - tmp;
            }

            long sum = gm * nk[1] % 1000000007;

            sum += cnt_loop % 1000000007 * (nk[1]*nk[1]-1)/2 % 1000000007 ;

            //for (int i = 0; i < nk[1]; i++)
            //{
            //    sum += (cnt_loop * i) % 1000000007;
            //}

            WriteLine(sum);
        }

        static int kaijou(int n)
        {
            int sum = 1;
            for (int i = 1; i <= n; i++)
            {
                sum = sum * i % 1000000007
;
            }
            return sum;
        }
    }
}