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.Text;
using System.Numerics;
using System.IO;

namespace SolutionCS
{
    class Program
    {
        private const long M = 1000000007;       // 素数(int範囲)

        static int N;
        static int K;

        static void Main(string[] args)
        {
            string[] s = Console.ReadLine().Split(' ');
            N = int.Parse(s[0]);
            K = int.Parse(s[1]);

            int[] v = new int[N];
            s = Console.ReadLine().Split(' ');
            for (int i = 0; i < N; i++)
            {
                v[i] = int.Parse(s[i]);
            }

            int left = 0;
            int right = 0;
            for (int i = 0; i < N; i++)
            {
                for (int j = i + 1; j < N; j++)
                {
                    if (v[i] < v[j])
                    {
                        left++;
                    }
                    else if(v[i] > v[j])
                    {
                        right++;
                    }
                }
            }

            long ans;

            checked
            {
                long m = K;
                m *= (K + 1);
                m /= 2;
                m %= M;

                long m0 = (K - 1);
                m0 *= K;
                m0 /= 2;
                m0 %= M;

                ans = right * m;
                ans %= M;

                ans += left * m0;
                ans %= M;
            }

            Console.WriteLine(ans);
        }
    }
}