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 Ruby (2.3.3) (judge verdict: Accepted)

str = gets.split
N,K = str[0].to_i, str[1].to_i
a = Array.new(N)
b = Array.new(2001,0)
c = 0
d = 0
mod = 10**9 + 7
str = gets.split
for i in 0...N do
    n = str[i].to_i
    a[i] = n
    b[n] += 1
end
for i in 1..2000 do
    b[i] += b[i-1]
end
for i in 1..2000 do
    d += (b[i]-b[i-1])*b[i-1]
end

x = 0
y = 1
while x < N-1
    y = x + 1
    while y < N
        if a[x] > a[y]
            c += 1
        end
        y += 1
    end
    x += 1
end

answer = (c * K) % mod
answer += (( K * (K-1) * d)/2)%mod
answer = answer % mod

puts answer