output_description: Print the amount of money that Iroha will hand to the cashier.
  
  * * *
input_description: The input is given from Standard Input in the following format:
  
      
      
      N K
      D_1 D_2 … D_K
attempt: import sys
  import math
  import collections
  import itertools
  import array
  import inspect
  
  # Set max recursion limit
  sys.setrecursionlimit(1000000)
  
  
  # Debug output
  def chkprint(*args):
      names = {id(v): k for k, v in inspect.currentframe().f_back.f_locals.items()}
      print(", ".join(names.get(id(arg), "???") + " = " + repr(arg) for arg in args))
  
  
  # Binary converter
  def to_bin(x):
      return bin(x)[2:]
  
  
  def li_input():
      return [int(_) for _ in input().split()]
  
  
  def gcd(n, m):
      if n % m == 0:
          return m
      else:
          return gcd(m, n % m)
  
  
  def gcd_list(L):
      v = L[0]
  
      for i in range(1, len(L)):
          v = gcd(v, L[i])
  
      return v
  
  
  def lcm(n, m):
      return (n * m) // gcd(n, m)
  
  
  def lcm_list(L):
      v = L[0]
  
      for i in range(1, len(L)):
          v = lcm(v, L[i])
  
      return v
  
  
  # Width First Search (+ Distance)
  def wfs_d(D, N, K):
      """
      D: 隣接行列(距離付き)
      N: ノード数
      K: 始点ノード
      """
  
      dfk = [-1] * (N + 1)
      dfk[K] = 0
  
      cps = [(K, 0)]
      r = [False] * (N + 1)
      r[K] = True
      while len(cps) != 0:
          n_cps = []
          for cp, cd in cps:
              for i, dfcp in enumerate(D[cp]):
                  if dfcp != -1 and not r[i]:
                      dfk[i] = cd + dfcp
                      n_cps.append((i, cd + dfcp))
                      r[i] = True
  
          cps = n_cps[:]
  
      return dfk
  
  
  # Depth First Search (+Distance)
  def dfs_d(v, pre, dist):
      """
      v:  現在のノード
      pre: １つ前のノード
      dist: 現在の距離
  
      以下は別途用意する
      D: 隣接リスト(行列ではない)
      D_dfs_d: dfs_d関数で用いる，始点ノードから見た距離リスト
      """
  
      global D
      global D_dfs_d
  
      D_dfs_d[v] = dist
  
      for next_v, d in D[v]:
          if next_v != pre:
              dfs_d(next_v, v, dist + d)
  
      return
  
  
  def sigma(N):
      ans = 0
      for i in range(1, N + 1):
          ans += i
      return ans
  
  
  def comb(n, r):
      if n - r < r:
          r = n - r
      if r == 0:
          return 1
      if r == 1:
          return n
  
      numerator = [n - r + k + 1 for k in range(r)]
      denominator = [k + 1 for k in range(r)]
  
      for p in range(2, r + 1):
          pivot = denominator[p - 1]
          if pivot > 1:
              offset = (n - r) % p
              for k in range(p - 1, r, p):
                  numerator[k - offset] /= pivot
                  denominator[k] /= pivot
  
      result = 1
      for k in range(r):
          if numerator[k] > 1:
              result *= int(numerator[k])
  
      return result
  
  
  def bisearch(L, target):
      low = 0
      high = len(L) - 1
  
      while low <= high:
          mid = (low + high) // 2
          guess = L[mid]
          if guess == target:
              return True
          elif guess < target:
              low = mid + 1
          elif guess > target:
              high = mid - 1
      if guess != target:
          return False
  
  
  # --------------------------------------------
  
  dp = None
  
  
  def main():
      N, K = li_input()
      D = input().split()
  
      c_money = N
  
      while True:
          is_dislike = False
  
          for a in set(str(c_money)):
              if a in D:
                  is_dislike = True
                  break
  
          if is_dislike:
              c_money += 1
          else:
              print(c_money)
              return
  
  
  main()
  
problem_description: Statement
  
  Iroha is very particular about numbers. There are K digits that she dislikes:
  D_1, D_2, ..., D_K.
  
  She is shopping, and now paying at the cashier. Her total is N yen (the
  currency of Japan), thus she has to hand at least N yen to the cashier (and
  possibly receive the change).
  
  However, as mentioned before, she is very particular about numbers. When she
  hands money to the cashier, the decimal notation of the amount must not
  contain any digits that she dislikes. Under this condition, she will hand the
  minimum amount of money.
  
  Find the amount of money that she will hand to the cashier.
samples: [{"input": "1000 8\n    1 3 4 5 6 7 8 9", "output": "2000\n    \n\nShe dislikes all digits except 0 and 2.\n\nThe smallest integer equal to or greater than N=1000 whose decimal notation\ncontains only 0 and 2, is 2000.\n\n* * *"}, {"input": "9999 1\n    0", "output": "9999"}]