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)

n, k = gets.split.map(&:to_i)
a = gets.split.map(&:to_i)
b = a + a
c = 0
(a.size - 1).times do |i|
  (i + 1).upto(a.size - 1) do |j|
    c += 1 if a[i] > a[j]
  end
end
d = 0
(b.size - 1).times do |i|
  (i + 1).upto(b.size - 1) do |j|
    d += 1 if b[i] > b[j]
  end
end
def modpow(n, p, mod)
  r = 1
  while p > 0
    r = r * n % mod if p & 1 == 1;
    n = n * n % mod
    p >>= 1
  end
  r
end
def modinv(n, mod)
  modpow(n, mod - 2, mod)
end
d -= c * 2
MOD = 1_000_000_007
ans = (k - 1) * d % MOD
ans += 2 * c
ans %= MOD
ans *= k
ans %= MOD
ans *= modinv(2, MOD)
ans %= MOD
puts ans