output_description: Print the inversion number of B, modulo 10^9 + 7.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N K
      A_0 A_1 ... A_{N - 1}
attempt: import logging
  
  logging.basicConfig(level=logging.INFO, format="%(message)s")
  # logging.disable(logging.CRITICAL)
  
  
  def count_inversion(sequence):
      return sum(sum(m < n for m in sequence[i + 1 :]) for i, n in enumerate(sequence))
  
  
  def count_ponpon(sequence):
      counter = 0
      split_count = 0
      splited = 0
      num = len(sequence)
      for i in sequence:
          split_count = sequence.count(i)
          counter += num - split_count - splited
          splited += 1
  
      logging.info(counter)
      return counter
  
  
  def main():
      N, K = map(int, input().split())
      A_1 = list(map(int, input().split()))
      MODI = 10**9 + 7
      logging.info("Hello!")
  
      A_counter = count_inversion(A_1)
      k_counter = int((K) * (K - 1) / 2)
  
      A_ = count_ponpon(A_1)
  
      print((A_counter * K + k_counter * A_) % MODI)
  
  
  if __name__ == "__main__":
      main()
  
problem_description: 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.
samples: [{"input": "2 2\n    2 1", "output": "3\n    \n\nIn this case, B~=~2,~1,~2,~1. We have:\n\n  * B_0 > B_1\n  * B_0 > B_3\n  * B_2 > B_3\n\nThus, the inversion number of B is 3.\n\n* * *"}, {"input": "3 5\n    1 1 1", "output": "0\n    \n\nA may contain multiple occurrences of the same number.\n\n* * *"}, {"input": "10 998244353\n    10 9 8 7 5 6 3 4 2 1", "output": "185297239\n    \n\nBe sure to print the output modulo 10^9 + 7."}]