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: Wrong Answer)

import java.util.Scanner;
public class Main {
	static final long MOD = 1000000007L;

	public static void main(String[] args){
		Main main = new Main();
		main.solveC();
	}

	private void solveA() {
		Scanner sc = new Scanner(System.in);
		int five = 0;
		int seven = 0;
		for (int i = 0; i < 3; i++) {
			switch (sc.nextInt()) {
			case 5:
				five++;
				break;
			case 7:
				seven++;
				break;
			default:
			}
		}
		if (five == 2 && seven == 1) {
			System.out.println("YES");
		} else {
			System.out.println("NO");
		}
	}

	private void solveC() {
		Scanner sc = new Scanner(System.in);
		String N = sc.next();
		int K = sc.nextInt();
		boolean[] forbid = new boolean[10];
		int[] next = new int[10];
		for (int i = 0; i < 10; i++) {
			forbid[i] = false;
			next[i] = i;
		}
		for (int i = 0; i < K; i++) {
			forbid[sc.nextInt()] = true;
		}
		for (int i = 0; i < 10; i++) {
			for (int j = 0; j < 10; j++) {
				if (!forbid[(i + j) % 10]) {
					next[i] = i + j;
					break;
				}
			}
		}

		int L = N.length();
		String answer = "";
		boolean lift = false;
		for (int i = 0; i < N.length(); i++) {
			char c = N.charAt(L - i - 1);
			int ctoi = c - '0' + (lift ? 1 : 0);
			int ni = next[ctoi % 10] % 10;
			if (ctoi == ni) {
				answer = ni + answer;
			} else {
				int AL = answer.length();
				answer = "" + ni;
				for (int j = 0; j < AL; j++) {
					answer = answer + next[0];
				}
			}
			lift = (ctoi >= 10 || next[ctoi] >= 10);
		}
		if (lift) {
			System.out.print("1");
		}
		System.out.println(answer);
	}

	private void solveD() {
		Scanner sc = new Scanner(System.in);
		long H = sc.nextLong();
		long W = sc.nextLong();
		long A = sc.nextLong();
		long B = sc.nextLong();

		long startRowIndex = H - A;
		long startColIndex = B + 1;
		long answer = 0L;
		for (long shiftIndex = 0; startRowIndex - shiftIndex > 0 && startColIndex + shiftIndex <= W; shiftIndex++) {
			answer += pointComb(H, W, H - A - shiftIndex, B + 1 + shiftIndex);
			answer %= MOD;
		}
		System.out.println(answer);
	}

	/**
	 * h行w列の格子について点(r, c)を通る組み合わせを求めます
	 * @param h 格子の行数
	 * @param w 格子の列数
	 * @param r 通る点の行index
	 * @param c 通る点の列index
	 * @return C(r + c - 2, c - 1) * C(h - r + w - c, w - c)
	 */
	private long pointComb(long h, long w, long r, long c) {
		return combination(r + c - 2, c - 1) * combination(h - r + w - c, w - c);
	}

	/**
	 * n個からm個選ぶ組み合わせの数を求めます
	 * @param n 全体の数
	 * @param m 選ぶ数
	 * @return C(n, m)
	 */
	private long combination(long n, long m) {
		if (n / 2 < m) {
			return combination(n , n - m);
		}
		if (m <= 0) {
			return 1;
		}
		return (combination(n, m - 1) * (n - m + 1) / m) % MOD;
	}
}