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 Julia (0.5.0) (judge verdict: Accepted)

parseInt(x) = parse(Int, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)

function main()
	mod = 10^9+7
	n,k = readline() |> split |> parseMap
	a = readline() |> split |> parseMap
	x = zeros(Int,n)
	y = zeros(Int,n)
	for i in 1:n
		for j in 1:n
			if j<i
				if a[i]>a[j]
					x[i]+=1
				end
			elseif i<j
				if a[i]>a[j]
					y[i]+=1
				end
			end
		end
	end
	z = 0
	for i in 1:n
		z = (z+(k*y[i])%mod+div(k*(k-1),2)%mod*(x[i]+y[i])%mod)%mod
	end
	println(z)
end

main()