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: a = input().split(" ")
  N = int(a[0])
  K = int(a[1])
  sequence_list = input().split(" ")
  sequence = []
  for i in range(0, N):
      sequence.append(int(sequence_list[i]))
  count = 0
  for p in range(0, N):
      count1 = 0
      count2 = 0
      for q in range(0, p):
          if sequence[p] - sequence[q] > 0:
              count1 += 1
              print(sequence[p])
              print("a")
      if p != N - 1:
          for r in range(p, N):
              if sequence[p] - sequence[r] > 0:
                  count2 += 1
                  print(sequence[p])
                  print("b")
          count += ((count1 + count2) * (K - 1) + count2 + count2) * K / 2
      else:
          count += (count1 * (K - 1)) * K / 2
          print(sequence[p])
          print("c")
  print(int(count % (10**9 + 7)))
  
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."}]