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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define MOD1 1000000007
#define MOD2 998244353
#define LIMIT1 200002
#define LIMIT2 500002
typedef long long ll;
typedef long double ld;
typedef const void cv;
#define rep(i,n) for((i)=0;(i)<(n);(i)++)
#define per(i,n) for((i)=(n)-1;(i)>=0;(i)--)
#define max(a,b) ((a)>(b) ? (a) : (b))
#define min(a,b) ((a)<(b) ? (a) : (b))
#define zt(a,b) (max((a),(b))-min((a),(b)))
#define absl(v) labs((v))
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll nPr(int n,int r){ll i,result=1;rep(i,r){result*=(ll)(n-i);}return result;}
ll nCr(int n,int r){ll i,result=1;if(n<r) return 0;rep(i,min(r,n-r)){result*=(ll)(n-i);result/=(ll)(i+1);}return result;}
#define nHr(n,r) nCr((int)((n)+(r)+1),(int)(r))
#define sankaku(x) (((x)*((x)+1))/2)
int dx[8]={1,0,-1,0,1,-1,-1,1},dy[8]={0,1,0,-1,1,1,-1,-1};
int upint(cv *a, cv *b) { return *(int *)a < *(int *)b ? -1 : *(int *)a > *(int *)b ? 1 : 0; }
int downint(cv *a, cv *b) { return *(int *)a < *(int *)b ? 1 : *(int *)a > *(int *)b ? -1 : 0; }
int upchar(cv *left, cv *right) {return strcmp((char *)left,(char *)right);}
int downchar(cv *left, cv *right) {return strcmp((char *)right,(char *)left);}
void tmpInit(int *c,ll m) {ll i;rep(i,m) c[i]=i;}
ll modpow(ll a, ll n, ll mod) {
  ll res = 1;
  while (n > 0) {
    if (n & 1){
      res *= a; if(mod>=1) res %= mod;
    }
    a *= a; if(mod>=1) a %= mod;
    n >>= 1;
  }
  return res;
}
void initialize(){
    
}
int a[LIMIT2]={0};
ll k;
int find(ll price){
  ll i,j;
  j=price%10;
  if(price<=0) return 1;
  rep(i,k) if(a[i]==j) return 0;
  return find(price/10);
}
int main(void){
  initialize();
  ll n,m,i,j,result=0;
  char s[LIMIT1];
  
  scanf("%lld %lld",&n,&k);
  rep(i,k) scanf("%d",&a[i]);
  i=n;
  while(1){
    if(find(i)==1){
      result = i;
      break;
    }
    i++;
  }
  printf("%lld\n",result);
  return 0;
}