output_description: Print the minimum number of times the magician needs to use the magic. If he
  cannot reach (D_h,D_w), print `-1` instead.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      H W
      C_h C_w
      D_h D_w
      S_{11}\ldots S_{1W}
      \vdots
      S_{H1}\ldots S_{HW}
attempt: 
  from collections import deque
  
  
  H, W = [int(x) for x in input().split()]
  Ch, Cw = [int(x) - 1 for x in input().split()]
  Dh, Dw = [int(x) - 1 for x in input().split()]
  
  warp_list = [(-2, -2), (-2, -1), (-2, 0), (-2, 1), (-2, 2),
               (-1, -2), (-1, -1), (-1, 1), (-1, 2),
               (0, -2), (0, 2),
               (1, -2), (1, -1), (1, 1), (1, 2),
               (2, -2), (2, -1), (2, 0), (2, 1), (2, 2)]
  
  maze = []
  warp_count = []
  for _ in range(H):
      line = list([x for x in input()])
      maze.append(line)
      warp_count.append([-1] * W)
  
  queue = deque()
  warp_queue = deque()
  
  queue.appendleft((Ch, Cw, 0))
  warp_count[Ch][Cw] = 0
  
  answer = -1
  while queue:
      def maze_check(h, w, use_magic):
          if h < 0 or H <= h:
              return False
          if w < 0 or W <= w:
              return False
          if maze[h][w] == '#':
              return False
          if warp_count[h][w] == -1:
              return True
          if warp_count[h][w] <= use_magic:
              return False
  
          return True
  
      now_h, now_w, use_magic = queue.pop()
  
      if now_h == Dh and now_w == Dw:
          if answer == -1:
              answer = use_magic
  
          elif answer > use_magic:
              answer = use_magic
  
          continue
  
      # 上下左右に移動
      do_warp = True
      if maze_check(now_h+1, now_w, use_magic):
          queue.appendleft((now_h + 1, now_w, use_magic))
          warp_count[now_h + 1][now_w] = use_magic
          do_warp = False
      if maze_check(now_h-1, now_w, use_magic):
          queue.appendleft((now_h - 1, now_w, use_magic))
          warp_count[now_h - 1][now_w] = use_magic
          do_warp = False
      if maze_check(now_h, now_w+1, use_magic):
          queue.appendleft((now_h, now_w + 1, use_magic))
          warp_count[now_h][now_w + 1] = use_magic
          do_warp = False
      if maze_check(now_h, now_w-1, use_magic):
          queue.appendleft((now_h, now_w - 1, use_magic))
          warp_count[now_h][now_w - 1] = use_magic
          do_warp = False
  
      if do_warp:
          warp_queue.appendleft((now_h, now_w, use_magic))
  
      if not queue:
          while warp_queue:
              now_h, now_w, use_magic = warp_queue.pop()
  
              for y, x in warp_list:
                  warp_h = now_h + y
                  warp_w = now_w + x
  
                  if maze_check(warp_h, warp_w, use_magic+1):
                      queue.append((warp_h, warp_w, use_magic+1))
                      warp_count[warp_h][warp_w] = use_magic + 1
  
  print(answer)
  
problem_description: Statement
  
  A maze is composed of a grid of H \times W squares - H vertical, W horizontal.
  
  The square at the i-th row from the top and the j-th column from the left -
  (i,j) \- is a wall if S_{ij} is `#` and a road if S_{ij} is `.`.
  
  There is a magician in (C_h,C_w). He can do the following two kinds of moves:
  
    * Move A: Walk to a road square that is vertically or horizontally adjacent to the square he is currently in.
    * Move B: Use magic to warp himself to a road square in the 5\times 5 area centered at the square he is currently in.
  
  In either case, he cannot go out of the maze.
  
  At least how many times does he need to use the magic to reach (D_h, D_w)?
samples: [{"input": "4 4\n    1 1\n    4 4\n    ..#.\n    ..#.\n    .#..\n    .#..", "output": "1\n    \n\nFor example, by walking to (2,2) and then using the magic to travel to (4,4),\njust one use of magic is enough.\n\nNote that he cannot walk diagonally.\n\n* * *"}, {"input": "4 4\n    1 4\n    4 1\n    .##.\n    ####\n    ####\n    .##.", "output": "-1\n    \n\nHe cannot move from there.\n\n* * *"}, {"input": "4 4\n    2 2\n    3 3\n    ....\n    ....\n    ....\n    ....", "output": "0\n    \n\nNo use of magic is needed.\n\n* * *"}, {"input": "4 5\n    1 2\n    2 5\n    #.###\n    ####.\n    #..##\n    #..##", "output": "2"}]