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)

#include <bits/stdc++.h>
using namespace std;
 
#define int long long
#define pb push_back
#define ub upper_bound
#define lb upper_bound
#define mp make_pair
#define F first
#define S second
#define FOR(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define REP(i,n) FOR(i,0,n)
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define SORT(c) sort((c).begin(),(c).end())
#define ve vector
#define vi vector<int>
#define vp vector<pair<int,int>>
#define vvi vector<vector<int>>
 
typedef long long ll;
const ll INF = LLONG_MAX - 100;
const ll mod = 1e9 + 7;
const int MAX_N = 2e5 + 5;
int dx[] = {-1,0,1,0}, dy[] = {0,1,0,-1};
vector<ll> prime;
 
int fac[MAX_N], inv[MAX_N];
 
template <class T = ll> T in() {T x; cin >> x; return (x);}
void DEBUG(vector<int> a) {for(int i=0;i<a.size();i++)cout<<a[i]<<" ";cout<<endl;}
void EMP(int x) {cout<<"!!!"<<x<<"!!!"<<endl;}
ll GCD(ll a, ll b) {ll c; while (b != 0) {c = a % b; a = b; b = c;}return a;}
ll LCM(ll a, ll b) {return a * b / GCD(a, b);}
ll POW(ll a, ll b, bool usemod = true) {ll c = 1LL; while (b > 0) {if (b & 1LL) {if(!usemod)c*=a;else c = a * c%mod;}if (!usemod) a*=a;else a = a * a%mod; b >>= 1LL;}return c;}
void _nCr() {fac[0] = 1LL; for (int i = 1LL; i < MAX_N; i++) {fac[i] = fac[i - 1LL] * i%mod;}for (int i = 0; i < MAX_N; i++) {inv[i] = POW(fac[i], mod - 2);}}
ll nCr(ll n, ll r) {return (fac[n] * inv[r] % mod)*inv[n - r] % mod;}
void PRI(ll n) {bool a[n + 1LL]; for (int i = 0; i < n + 1LL; i++) {a[i] = 1LL;}for (int i = 2; i < n + 1LL; i++) {if (a[i]) {prime.pb(i); ll b = i; while (b <= n) {a[b] = 0; b += i;}}}}
template <typename T> T chmin(T& a, T b) {if(a>b)a=b;return a;}
template <typename T> T chmax(T& a, T b) {if(a<b)a=b;return b;}

class Warshall_Floyd {
public:
	vector<vector<int>> d;
	int nodecnt = -1;

	Warshall_Floyd(int n) {
		d = vector<vector<int>>(n,vector<int>(n,INF));
		nodecnt = n;
	}

	void add(int s,int e,int cost) {
		d[s][e] = cost;
	}

	bool wf() {
		for (int k = 0;k < nodecnt;k++) {
			for (int i = 0;i < nodecnt;i++) {
				for (int j = 0;j < nodecnt;j++) {
					//k経由でi->j
					d[i][j] = min(d[i][j],d[i][k] + d[k][j]);
				}
			}
		}
	}
};

bool ng[10];
bool check(int n) {
	int keta = 1;
	while (true) {
		if (pow(10,keta) > n) break;
		keta++;
	}
	REP (i,keta) {
		if (ng[n % 10]) {return false;}
		n /= 10;
	}
	return true;
}

bool solve() {
	int n,k; cin >> n >> k;
	REP (i,k) {int x;cin >> x; ng[x] = true;}
	int tmp = n;
	while (true) {
		if (check(tmp)) {cout << tmp << endl; return 0;}
		tmp++;
	}
	// cout << check(tmp) << endl;
}

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	solve();
}