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 Python (3.4.3) (judge verdict: Wrong Answer)

from decimal import *
getcontext().prec = 50
N, K = map(int, input().split())
A = list(map(int, input().split()))

x = 0
for l in range(N - 1):
  for r in range(l + 1, N):
    if A[l] > A[r]:
      x += 1
# print(x)
      
AA = A + A
y = 0
for l in range(2 * N - 1):
  for r in range(l + 1, 2 * N):
    if AA[l] > AA[r]:
      y += 1
y -= 2 * x
# print(y)

ans = Decimal(K * x % (10 ** 9 + 7)) + Decimal(K * (K - 1) * y / 2 % (10 ** 9 + 7))
ans = Decimal(ans % (10 ** 9 + 7))
print(int(ans))
# print(int((K * x  % (10 ** 9 + 7) + K * (K - 1) * y / 2  % (10 ** 9 + 7)) % (10 ** 9 + 7)))
#kkmod = (K * (K - 1)) % (10 ** 9 + 7)
#Kxmod = (K * x) % (10 ** 9 + 7)
#ans = (2 * Kxmod +  kkmod * y) % (10 ** 9 + 7)
#print(int(ans / 2 % (10 ** 9 + 7)))