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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;


/// File input & output /////
#define fRead(x)        freopen("x.txt","r",stdin)
#define fWrite(x)       freopen ("x.txt","w",stdout)


#define ld              long double
#define ll              long long
#define ull             unsigned long long
#define ff              first
#define ss              second
#define pb              push_back
#define INF             2e16
#define PI              acos(-1.0)
#define mp              make_pair
#define pii             pair<int,int>
#define pll             pair<LL,LL>


#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(a,min(b,min(c,d)))
#define max4(a,b,c,d)   max(a,max(b,max(c,d)))
#define SQR(a)          ((a)*(a))
#define FOR(i,a,b)      for(int i=a;i<=b;i++)
#define ROF(i,a,b)      for(int i=a;i>=b;i--)
#define REP(i,b)        for(int i=0;i<b;i++)
#define MEM(a,x)        memset(a,x,sizeof(a))
#define ABS(x)          ((x)<0?-(x):(x))

#define SORT(v)         sort(v.begin(),v.end())
#define REV(v)          reverse(v.begin(),v.end())


#define FASTIO        ios_base::sync_with_stdio(0);cin.tie(nullptr);




///..................Sort by second element in a pair in decending order............. ///
bool SortbySecDesc(const pair<long long,long long > &a, const pair<long long,long long > &b)
{
    return a.second>b.second;
}

///.........................first element of pair in descending order..........................//////
bool sortinrev(const pair<int,int> &a,
               const pair<int,int> &b)
{
       return (a.first > b.first);
}



///.............................................................TEMPLATE................................................./////

int main()
{
    FASTIO;

    int k;

     int ar[100];

    memset(ar,-1,sizeof(ar));


    string num;

    cin >> num >> k;







   // cout << num << endl;


    for(int i=0; i<k; i++)
    {

        int b;

        cin >> b;

        ar[b]++;
    }

    //sort(ar,ar+k);

    vector<int>v;

    for(int i=0; i<=9; i++)
    {
        if(ar[i]==-1)
        {
            v.pb(i);
           // cout << i << endl;

        }
    }

    sort(v.begin(),v.end());



    for(int i=0; i<num.length(); i++)
    {
        int a = (int)(num[i]-48);

        //cout << a << endl;

        if(ar[a]!=-1)
        {
            if(i==0)
            {
                for(int j=0; j<v.size(); j++)
                {
                    if(v[j]!=0)
                    {
                       // cout << v[j] << endl;

                        num[i] = v[j]+'0';
                        break;
                    }
                }

            }
            else
            {
                num[i] = v[0]+'0';
            }

        }

    }

    cout << num << endl;





    return 0;
}