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
		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 = scan.next();
			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).anyMatch(dd -> dd == i)).toArray();
		
		// 各位の数字の取得
		final int[] value = new int[n.length()];		// 1≦N<10000より4桁
		for (int i = 0; i < value.length; i++) {
			value[i] = Integer.parseInt(n.substring(i, i + 1));
		}
		
		int[] result = new int[n.length() + 1];
		boolean big = false;	//大きいことが確定すればtrue
		
		result[0] = 0;
		for (int j = 0; j < n.length(); j++) {
			final int i = j;

			if (big) {
				result[i + 1] = like[0];
			} else {
				OptionalInt temp = Arrays.stream(like).filter(likevalue -> likevalue >= value[i]).min();
				
				if (temp.isPresent()) {
					result[i + 1] = temp.getAsInt();
					big = value[i] < result[i + 1];
				} else {			// value[i] > likeMax
					moveUp(result, like, i);
					big = true;
				}
			}
						
		}

		
		//数値に戻す
		// 出力
		out.println(Arrays.stream(result).reduce((start, element) -> start * 10 + element).getAsInt());
		out.flush();
	}
	
	/** 繰り上がり */
	static void moveUp(int[] result, int[] like, int i) {
		OptionalInt temp = Arrays.stream(like).filter(likevalue -> likevalue > result[i]).min();
		result[i+1] = like[0];
		if (temp.isPresent()) {
			result[i] = temp.getAsInt();		// i == 0でlike[0]==0 ? like[1] : like[0]になる
		} else {								// よって、ここでは i > 1
			moveUp(result, like, i -1);
		}
	}
	
}