output_description: Output should be in the following format:
  
    * In the first line, print integers h and w representing the size of the grid you constructed, with a space in between.
    * Then, print h more lines. The i-th (1 \leq i \leq h) of these lines should contain a string s_i as follows:
      * If the square at the i-th row and j-th column (1 \leq j \leq w) in the grid is painted white, the j-th character in s_i should be `.`.
      * If the square at the i-th row and j-th column (1 \leq j \leq w) in the grid is painted black, the j-th character in s_i should be `#`.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      A B
attempt: ###############################################################################
  
  from sys import stdout
  from bisect import bisect_left as binl
  from copy import copy, deepcopy
  from collections import defaultdict
  
  
  mod = 1
  
  
  def intin():
      input_tuple = input().split()
      if len(input_tuple) <= 1:
          return int(input_tuple[0])
      return tuple(map(int, input_tuple))
  
  
  def intina():
      return [int(i) for i in input().split()]
  
  
  def intinl(count):
      return [intin() for _ in range(count)]
  
  
  def modadd(x, y):
      global mod
      return (x + y) % mod
  
  
  def modmlt(x, y):
      global mod
      return (x * y) % mod
  
  
  def lcm(x, y):
      while y != 0:
          z = x % y
          x = y
          y = z
      return x
  
  
  def combination(x, y):
      assert x >= y
  
      if y > x // 2:
          y = x - y
  
      ret = 1
  
      for i in range(0, y):
          j = x - i
          i = i + 1
  
          ret = ret * j
          ret = ret // i
  
      return ret
  
  
  def get_divisors(x):
      retlist = []
      for i in range(1, int(x**0.5) + 3):
          if x % i == 0:
              retlist.append(i)
              retlist.append(x // i)
      return retlist
  
  
  def get_factors(x):
      retlist = []
      for i in range(2, int(x**0.5) + 3):
          while x % i == 0:
              retlist.append(i)
              x = x // i
      retlist.append(x)
      return retlist
  
  
  def make_linklist(xylist):
      linklist = {}
      for a, b in xylist:
          linklist.setdefault(a, [])
          linklist.setdefault(b, [])
          linklist[a].append(b)
          linklist[b].append(a)
      return linklist
  
  
  def calc_longest_distance(linklist, v=1):
      distance_list = {}
      distance_count = 0
      distance = 0
      vlist_previous = []
      vlist = [v]
      nodecount = len(linklist)
  
      while distance_count < nodecount:
          vlist_next = []
          for v in vlist:
              distance_list[v] = distance
              distance_count += 1
              vlist_next.extend(linklist[v])
          distance += 1
          vlist_to_del = vlist_previous
          vlist_previous = vlist
          vlist = list(set(vlist_next) - set(vlist_to_del))
  
      max_distance = -1
      max_v = None
      for v, distance in distance_list.items():
          if distance > max_distance:
              max_distance = distance
              max_v = v
  
      return (max_distance, max_v)
  
  
  def calc_tree_diameter(linklist, v=1):
      _, u = calc_longest_distance(linklist, v)
      distance, _ = calc_longest_distance(linklist, u)
      return distance
  
  
  class UnionFind:
  
      def __init__(self, n):
          self.parent = [i for i in range(n)]
  
      def root(self, i):
          if self.parent[i] == i:
              return i
          self.parent[i] = self.root(self.parent[i])
          return self.parent[i]
  
      def unite(self, i, j):
          rooti = self.root(i)
          rootj = self.root(j)
          if rooti == rootj:
              return
          if rooti < rootj:
              self.parent[rootj] = rooti
          else:
              self.parent[rooti] = rootj
  
      def same(self, i, j):
          return self.root(i) == self.root(j)
  
  
  ###############################################################################
  
  
  def main():
      a, b = intin()
  
      if a < b:
          dot = "#"
          wide = "."
      else:
          dot = "."
          wide = "#"
  
      print("38 100")
  
      print(dot * 100)
      print(("%s%s" % (dot, wide)) * 50)
  
      mincount = min(a, b) - 1
  
      for i in range(12):
          lline = ""
          for j in range(50):
              if mincount == 0:
                  lline += "%s%s" % (wide, wide)
              elif i & 1:
                  lline += "%s%s" % (dot, wide)
                  mincount -= 1
              else:
                  lline += "%s%s" % (wide, dot)
                  mincount -= 1
          print(lline)
  
      dotcount = abs(a - b)
  
      for i in range(12):
          print(wide * 100)
          lline = ""
          for j in range(50):
              if dotcount == 0:
                  lline += "%s%s" % (wide, wide)
              else:
                  lline += "%s%s" % (wide, dot)
                  dotcount -= 1
          print(lline)
  
  
  if __name__ == "__main__":
      main()
  
problem_description: Statement
  
  You are given two integers A and B.
  
  Print a grid where each square is painted white or black that satisfies the
  following conditions, in the format specified in Output section:
  
    * Let the size of the grid be h \times w (h vertical, w horizontal). Both h and w are at most 100.
    * The set of the squares painted white is divided into exactly A connected components.
    * The set of the squares painted black is divided into exactly B connected components.
  
  It can be proved that there always exist one or more solutions under the
  conditions specified in Constraints section. If there are multiple solutions,
  any of them may be printed.
samples: [{"input": "2 3", "output": "3 3\n    ##.\n    ..#\n    #.#\n    \n\nThis output corresponds to the grid below:\n\n![2701558bf42f7c088abad927b419472a.png](https://img.atcoder.jp/arc093/2701558bf42f7c088abad927b419472a.png)\n\n* * *"}, {"input": "7 8", "output": "3 5\n    #.#.#\n    .#.#.\n    #.#.#\n    \n\n* * *"}, {"input": "1 1", "output": "4 2\n    ..\n    #.\n    ##\n    ##\n    \n\n* * *"}, {"input": "3 14", "output": "8 18\n    ..................\n    ..................\n    ....##.......####.\n    ....#.#.....#.....\n    ...#...#....#.....\n    ..#.###.#...#.....\n    .#.......#..#.....\n    #.........#..####."}]