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: Runtime Error)

n, k = map(int, input().split())
a = tuple(map(int, input().split()))
d = 10**9+7

class Bit:
  def __init__(self, n):
    self.size = n
    self.tree = [0] * (n + 1)
 
  def sum(self, i):
    s = 0
    while i > 0:
      s += self.tree[i]
      i -= i & -i
    return s
 
  def add(self, i, x):
    while i <= self.size:
      self.tree[i] += x
      i += i & -i
       
bit = Bit(n)
rbit = Bit(n)
left = []
right = []
for i, (p, rp) in enumerate(zip(a, reversed(a))):
  bit.add(p, 1)
  rbit.add(rp, 1)
  left.append(i + 1 - bit.sum(p))
  right.append(i + 1 - rbit.sum(rp))

left = sum(left)
right = sum(right)

def ssum(a, d, n):
  return n*(2*a+(n-1)*d)//2

print(
  (
    ssum(left, left, k)
    + ssum(0, right, k)
  )%d
)