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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

//各桁の数字と桁数を求める関数
int to10Digit(int n, int b[]);

int main(){
	int b[6]={0};	//10進数各桁の数字を格納する配列
	int d[10]={0};	//嫌いな数値の時1となるフラグ
	int n;			//物を買うのに必要な金額
	int k;			//嫌いな数字の個数
	int i;			//for文などで用いるカウンタ
	int num;		//嫌いな数字を一時的に格納する変数
	int a;			//nの10進桁数

	//金額と嫌いな数値の個数入力
	scanf("%d %d",&n,&k);
	for(i=0;i<k;i++){
		scanf("%d",&num);
		d[num]=1;
	}
	//払うべき金額が0の時0を出力
	if(n==0){
		printf("0\n");
		return 0;
	}

	//各桁の数値と桁数を求める
	a=to10Digit(n,b);

	//嫌いな数字でなくなるまで加算
	i=0;
	while(i<a){
		//ある桁が10以上となったとき次の桁加算
		if(b[i]>9){
			//桁上がりの例外処理
			if(i+1>a-1)
				a++;
			b[i+1]++;
			b[i]=0;
		}
		//嫌いな数字であれば桁数値bを加算
		if(d[b[i]]!=0)
			b[i]++;
		//嫌いな数字でなければ次の桁へ
		else
			i++;
	}
	//払う金額を表示
	for(i=0;i<a;i++)
		printf("%d",b[a-i-1]);
	printf("\n");

	return 0;
}

int to10Digit(int n,int b[]){
	int i=0;
	while(n>0){
		b[i]=n%10;
		n/=10;
		i++;
	}
	return i;
}