output_description: Print the minimum number of operations needed.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      H W
      s_{11} s_{12} \cdots s_{1W}
      s_{21} s_{22} \cdots s_{2W}
       \vdots
      s_{H1} s_{H2} \cdots s_{HW}
      
  
  Here s_{rc} represents the color of (r, c) \- `#` stands for black, and `.`
  stands for white.
attempt: H, W = map(int, input().split())
  
  S = [input() for _ in range(W)]
  
  M = []
  
  
  def dfs(m, n, history):
      history_ = history + S[m][n]
      if H - m > 1:
          dfs(m + 1, n, history_)
  
      if W - n > 1:
          dfs(m, n + 1, history_)
  
      if H - m == 1 and H - n == 1:
          count = 0
          for h in history_:
              if h == "#":
                  count += 1
          M.append(count)
  
      return
  
  
  dfs(0, 0, "")
  print(min(M))
  
problem_description: Statement
  
  Consider a grid with H rows and W columns of squares. Let (r, c) denote the
  square at the r-th row from the top and the c-th column from the left. Each
  square is painted black or white.
  
  The grid is said to be _good_ if and only if the following condition is
  satisfied:
  
    * From (1, 1), we can reach (H, W) by moving one square **right or down** repeatedly, while always being on a white square.
  
  Note that (1, 1) and (H, W) must be white if the grid is good.
  
  Your task is to make the grid good by repeating the operation below. Find the
  minimum number of operations needed to complete the task. It can be proved
  that you can always complete the task in a finite number of operations.
  
    * Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) \- that is, from white to black and vice versa.
samples: [{"input": "3 3\n    .##\n    .#.\n    ##.", "output": "1\n    \n\nDo the operation with (r_0, c_0, r_1, c_1) = (2, 2, 2, 2) to change just the\ncolor of (2, 2), and we are done.\n\n* * *"}, {"input": "2 2\n    #.\n    .#", "output": "2\n    \n\n* * *"}, {"input": "4 4\n    ..##\n    #...\n    ###.\n    ###.", "output": "0\n    \n\nNo operation may be needed.\n\n* * *"}, {"input": "5 5\n    .#.#.\n    #.#.#\n    .#.#.\n    #.#.#\n    .#.#.", "output": "4"}]