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++14 (GCC 5.4.1) (judge verdict: Accepted)

/*                                       @@@******ANUSHK******@@@           
                                        ******@nu$hk**@nu$hk******                                                                                                                                                                                                                                                                                                                                                                (Hello)//JSM//*/      
#include<iostream>                     // @nu$hk          // @nu$hk
#include<vector>                      // @nu$hk            // @nu$hk
#include<algorithm>                  // @nu$hk              // @nu$hk 
#include<string.h>                  // @nu$hk                // @nu$hk
#include<bits/stdc++.h>            // @nu$hk****/*@nu$hk*\****// @nu$hk
#include<map>                     // @nu$hk                    // @nu$hk 
#include<cmath>
#include<cstring>                  // @nu$hk                      // @nu$hk
using namespace std;            // @nu$hk                        // @nu$hk
#define ll long long int       // @nu$hk                          // @nu$hk 
#define ld long double
#define MOD 1000000007 
#define PI 3.1415926535897932384626433832795                                               
#define YY cout<<"YES"
#define NN cout<<"NO"
#define EE cout<<"\n"
#define ne cout<<"-1"
#define pb push_back 
#define ff first
#define ss second
#define mkp make_pair    
#define mkt make_tuple
#define alsrt(v ) sort(v.begin(), v.end());      
#define resrt(v ) sort(v.rbegin(), v.rend());
#define sortstr(v) sort(v.begin(), v.end(), cmp) 
#define fo(i , n) for(ll i = 0 ; i < n ; i++)
#define vi vector<ll> 
#define pr pair<ll,ll>
#define tpl tuple<ll,ll,ll>
bool sortbysec(const pair<int,int> &a,const pair<int,int> &b) {return (a.second < b.second); }
ll containsDigit(ll number, ll digit){while (number != 0){int curr_digit = number % 10;if (curr_digit == digit) return 1;number /= 10;}return 0;}
ll soltn(ll x, ll n) {if(x>=0) return x%n;    else return n+x%n;}
ll bin_srch(ll L[], ll N,ll item){ll flag=-2;int l=0,u=N-1,m;while(l<=u) {m=(l+u)/2;if(item==L[m]) {flag=m-1; break;} else if(item>L[m]) l=m-1; else u=m+1;} return flag;}
ll modExp(ll b, ll p){ll res = 1;while(p > 0) {if(p & 1){res = (res * b) % MOD;}b = (b * b) % MOD;p >>= 1;}return res;}
ll properdiv(ll x,ll y){if(x%y==0) return x/y;else return x/y+1;}
ll coda(ll n,ll d){ll count=0;while(n!=0){if(n%10==d)count++;n/=10;}return count;}
ll minDissame(ll a[], ll n)  {map<ll,ll> hmap; ll minDistance = INT_MAX;  ll previousIndex = 0, currentIndex = 0;  for (ll i = 0; i < n; i++) {  if (hmap.find(a[i])!=hmap.end()) {  currentIndex = i;     previousIndex = hmap[a[i]];   minDistance = min((currentIndex -        previousIndex),minDistance);        }hmap[a[i]] = i;  }    return (minDistance == INT_MAX ? -1 : minDistance);  } 
ll minDistanytwo(ll arr[], ll n, ll x, ll y) {ll p = -1, min_dist = INT_MAX; for(ll i=0 ; i<n ; i++) {if(arr[i]==x || arr[i]==y) {if( p != -1 && arr[i] != arr[p])  min_dist = min(min_dist , i-p); p=i;  }} if(min_dist==INT_MAX) return -1; return min_dist; } 
ll mdls(ll x){return max(x,-x);}
ll pow1(ll n,ll p){if(p==0)return 1;ll x=pow1(n, p/2);x=(x*x)%MOD;if(p%2==0) return x;else return (x*n)%MOD;}
string fumn(string s){string p;for(ll i = 0; i < s.length(); i++)if(isalpha(s[i]))p+=toupper(s[i]);return p;}
ll turns(string S,string SS){ ll i,cc=0,l=S.length();fo(i,l){ if(S[i]!=SS[i]) cc++; } return cc/2;}
ld polygonArea(ld X[], ld Y[], ll n) {  double area = 0.0; ll j = n - 1; for (ll i = 0; i < n; i++) {   area += (X[j] + X[i]) * (Y[j] - Y[i]);  j = i;  }  return abs(area / 2.0); } 
ll isSubstr(string s1, string s2) {   ll M = s1.length();    ll N = s2.length();  for (ll i = 0; i <= N - M; i++) { ll j; for (j = 0; j < M; j++) if (s2[i + j] != s1[j])  break;         if (j == M) return i;   } return -1; } 
//s2>s1;if(not) rtrn -1;
ll hhh2(unsigned ll n) {   if (n < 1)  return 0; ll res = 1;  ll u=0; for (ll i=0; i<8*sizeof(unsigned ll); i++) {    ll curr = 1 << i; if (curr > n)      break;   u++;  res = curr;} return u; }
ld ldgcd(ld a,ld b){ return a<0.0001?b:ldgcd(fmod(b,a),a);} 
void onesol(ll a,ll b,ll& x,ll& y,ll& g)
{
    if(!b) {g=a;x=1;y=0;}
    else{onesol(b,a%b,y,x,g);y-=a/b*x;}
}
/*struct pp
{
	ll x,y;
};*/
/*PLEASE START @ANU$HK GAUTAM
KEEP PATIENCE AND CONFIDENCE*/
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);		  ll t=1;
      
//	   cin>>t;
      
	      while(t--)
{
/*
1000 8
1 3 4 5 6 7 8 9
*/
ll n,k,i,j,l;
cin>>n>>k;
ll a[n];
map<ll,ll>mp;
fo(i,k)
{
	cin>>l;
	mp[l]++;
}
for(i=n;i<1000101001;i++)
{
	ll M=i,g=1;
	while(M!=0)
	{
		if(mp[M%10]!=0)
		{
			g=-1;
			break;
		}
		M/=10;
	}
	if(g==1)
	{
	cout<<i;
	break;
	}
	
}
}
return 0;
}