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: Runtime Error)

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]);

            string[] i = Console.ReadLine().Split(' ');
            int[] a = new int[N];
            
            for (int x = 0; x <= N-1; x++)
            {
                a[x] = int.Parse(i[x]);
            }

            int[] B = new int[N * K];
            for (int y = 0; y <=N*K-1; y++)
            {
                B[y] = a[y % N];
            }
            int count = 0;

            for (int u = 0; u <= N * K - 1; u++)
            {
                for(int v = 0; v <= N * K - 1; v++)
                {
                    if (u<v &&B[u]>B[v])
                    {
                        count++;
                    }
                }
            }
            count = count % 1000000007;

            Console.WriteLine(count);


        }

    }
}