Problem:
Score : 300 points

Problem 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.

Constraints

1 ≦ N < 10000

1 ≦ K < 10

0 ≦ D_1 < D_2 < … < D_K≦9

\{D_1,D_2,...,D_K\} ≠ \{1,2,3,4,5,6,7,8,9\}

Input

The input is given from Standard Input in the following format:

N K
D_1 D_2 … D_K

Output

Print the amount of money that Iroha will hand to the cashier.

Sample Input 1

1000 8
1 3 4 5 6 7 8 9

Sample Output 1

2000

She dislikes all digits except 0 and 2.

The smallest integer equal to or greater than N=1000 whose decimal notation contains only 0 and 2, is 2000.

Sample Input 2

9999 1
0

Sample Output 2

9999

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

N, K = gets.split.map(&:to_i)
hate_numbers = gets.split.map(&:to_i)
price = N.to_s.split("").map(&:to_i).reverse
payment = []
carrying = false # 繰り上がりが必要か否かを示すフラグ
price.each do |current_num|
  if carrying # 前の桁で繰り上がりが発生した場合
    current_num += 1 # 繰り上がりで1加算する
    if current_num >= 10
      carrying = true
      current_num %= 10
    end
  end

  while hate_numbers.include?(current_num)
    current_num += 1
    if current_num >= 10
      carrying = true
      current_num %= 10
    end
  end
  payment << current_num
end

# 一番大きな位まで処理したが、繰り上がりが発生している可能性があるためその処理
if carrying
  current_num = 1
  while hate_numbers.include?(current_num)
    current_num += 1
  end
  payment << current_num
end

# 結果の出力
puts payment.reverse.join("")