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: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

public class MainClass
{
	public const long Giri = 1000000007;

	
	public static void Main(string[] args)
	{
		string[] s = Input();
		int n = int.Parse(s[0]);
		long k = long.Parse(s[1]);
		int[] avs = new int[n];
		
		s = Input();
		for(int i = 0; i < n; i++)
		{
			avs[i] = int.Parse(s[i]);
		}
		long r = 0;
		long ol = 0;
		long ot = 0;
		for(int i = 0; i < n; i++)
		{
			long t = 0;
			for(int j = 0; j < i; j++)
			{
				if(avs[i]>avs[j])
					t++;
			}
			
			long l = 0;
			for(int j = i+1; j < n; j++)
			{
				if(avs[i]>avs[j])
					l++;
			}
			/*
			for(long q = k; q>=1; q--)
				r += (l * q + t * (q-1));
				*/
			ol += l;
			ot += t;
			r %= Giri;
		}
		
		for(long q = k; q>=1; q--)
			r += (ol * q + ot * (q-1)) % Giri;
		
		Console.WriteLine(r% Giri);
	}
	
	
	public static long GCD(long a, long b)
	{
		if(b == 1)
			return a;
		long p = b / a;
		return GCD(a, b - a*p);
	}
	
	public static string[] Input()
	{
		return Console.ReadLine().Split(' ');
	}

	
	public static int Clamp(int val, int max, int min = 0)
	{
		return Math.Min(max, Math.Max(min, val));
	}

}