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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.PrintWriter;
import java.util.Arrays;
import java.util.OptionalInt;
import java.util.Scanner;
import java.util.stream.IntStream;

public class Main {
	static PrintWriter out = new PrintWriter(System.out);
	public static void main(String[] args) {
		// 入力
		final String n;		// 1≦N<10000
		final int k;	// 1≦K<10
		final int[] d;	// 0≦D1<D2<…<DK≦9  {D1,D2,…,DK}≠{1,2,3,4,5,6,7,8,9}
		try(Scanner scan = new Scanner(System.in)) {
			n = "0" + scan.next();				// 繰り上がり用に先頭に0を挿入。
			k = scan.nextInt();
			d = new int[k];
			for (int i = 0; i < k; i++) {
				d[i] = scan.nextInt();
			}
		}
		
		// 嫌いでない数字の配列  maxはlike[10 - k -1] minはlike[0]
		final int[] like = IntStream.range(0, 10).filter(i -> Arrays.stream(d).noneMatch(dd -> dd == i)).toArray();
		
		// 各位の数字の取得		上記よりvalue[0] = 0が挿入されている。
		final int[] original = n.chars().map(i -> i - (int) '0').toArray();
		int[] result = Arrays.copyOf(original, original.length);
		
		// 末尾の桁から処理する。最上位の桁は別処理。
		for (int j = result.length - 1; j > 0; j--) {
			final int i = j;
			
			OptionalInt tempInt = Arrays.stream(like).filter(likevalue -> likevalue >= result[i]).min();
			if (tempInt.isPresent()) {
				result[i] = tempInt.getAsInt();
			} else {
				result[i - 1]++;	// 繰り上がり処理
				result[i] = like[0];
			}		
		}
		
		// 一番上の桁はlikeに0がなくても0でもよい。
		if (result[0] == 0) {
			result[0] = 0;
		} else {		// result[0] ==1で {D1,D2,…,DK}≠{1,2,3,4,5,6,7,8,9}により値があることが保証されている。
			result[0] = Arrays.stream(like).filter(likevalue -> likevalue >= result[0]).min().getAsInt();
		}
		
		// 上の方で値が増えれていれば、そこから先は関係ない。
		for (int i = 0; i < result.length - 1; i++) {
			if (result[i] > original[i]) {
				Arrays.fill(result, i + 1, result.length, like[0]);
				break;
			}
		}
		//上の桁から処理するよりも処理量が多くなってしまった。
		//まだ間違いがあるらしい。

		//数値に戻して出力
		out.println(Arrays.stream(result).reduce((start, element) -> start * 10 + element).getAsInt());
		out.flush();
	}
}