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;
using static System.Console;
using static System.Math;

class Program
{
    static void Main()
    {

        
        var NK = ReadLine().Split().Select(long.Parse).ToArray();
        var N = NK[0];
        var K = NK[1];
        var A = ReadLine().Split().Select(int.Parse).ToArray();
#if tru
        var B = new int[K*N];
        for (int i = 0; i < B.Length; i++)
        {
            B[i] = A[i % N];
        }
        var ans = 0;
        for (int i = 0; i < B.Length; i++)
        {
            for (int j = 0; j < B.Length; j++)
            {
                if (B[i] > B[j]) ans++;
            }
        }
        WriteLine(ans);
#else
        var mod = 1000000007L;
        var ans = 0L;
        var k = K * (K + 1) / 2 % mod;
        var t = K * (K - 1) / 2 % mod;
        //Console.WriteLine(k);
        for (int i = 0; i < N; i++)
        {
            for (int j = i + 1; j < N; j++)
            {
                if (A[i] > A[j])
                {
                    ans = (ans + k) % mod;
                }
                if(A[i]<A[j])
                {
                    ans = (ans + t) % mod;
                }
            }
        }
        WriteLine(ans);
#endif
    }
}