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

namespace AtTest.JSC_2019_qual
{
    class B
    {
        static void Main(string[] args)
        {
            var sw = new System.IO.StreamWriter(OpenStandardOutput()) { AutoFlush = false };
            SetOut(sw);

            Method(args);

            Out.Flush();
        }

        static void Method(string[] args)
        {
            int[] nk = ReadInts();
            long n = nk[0];
            long k = nk[1];
            long[] array = ReadLongs();
            long[] cnts = new long[2005];
            long[] smalls = new long[n];
            for(long i = n - 1; i >= 0; i--)
            {
                for(long j = 0; j < array[i]; j++)
                {
                    smalls[i] += cnts[j];
                }
                cnts[array[i]]++;
            }
            long[] allSmalls = new long[n];
            for(int i = 0; i < n; i++)
            {
                for(long j = 0; j < array[i]; j++)
                {
                    allSmalls[i] += cnts[j];
                }
            }
            long mask = 1000000000 + 7;
            long res = 0;
            for(int i = 0; i < n; i++)
            {
                res += smalls[i] * k;
                long tmp = (k - 1) * k / 2;
                tmp %= mask;
                tmp *= allSmalls[i];
                res += tmp;
                res %= mask;
            }
            WriteLine(res);
        }

        private static string Read() { return ReadLine(); }
        private static char[] ReadChars() { return Array.ConvertAll(Read().Split(), a => a[0]); }
        private static int ReadInt() { return int.Parse(Read()); }
        private static long ReadLong() { return long.Parse(Read()); }
        private static double ReadDouble() { return double.Parse(Read()); }
        private static int[] ReadInts() { return Array.ConvertAll(Read().Split(), int.Parse); }
        private static long[] ReadLongs() { return Array.ConvertAll(Read().Split(), long.Parse); }
        private static double[] ReadDoubles() { return Array.ConvertAll(Read().Split(), double.Parse); }
    }
}