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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <stdio.h>
#pragma optimize("O3")
#define sync                    \
ios_base::sync_with_stdio(0); \
cin.tie(0);                   \
cout.tie(0)
#define FOR(i,a,b) for (int i = (a); i < (b); i++)
#define RFOR(i,a,b) for (int i = (a - 1); i >= (b); i--)
#define all(a) (a).begin(), (a).end()
#define endl "\n"
#define ll long long
#define pb push_back
#define PII pair<int, int>
#define PLL pair<long long, long long>
#define vi vector<int>
#define vb vector<bool>
#define vc vector<char>
#define vll vector<long long>
#define mii map<int, int>
#define mll map<long long, long long>
#define MOD 1000000007
#define MAX 1000000007
#define MOD2 1234567891
#define SZ(a) (int) a.size()
#define ALL(a) a.begin(), a.end()
#define PB push_back
#define MP make_pair
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define TIE(l, r) Node *l, *r; tie(l, r)
#define EPS 0.00000001
#define PI 3.14159265359
void dout() { cerr << endl; }
#define MAXN 100001

template <typename Head, typename... Tail>
void dout(Head H, Tail... T) {
    cerr << H << ' ';
    dout(T...);
}

/*,--. ,--.
  |  .'   /  ,--,--.  ,---.   ,---.   ,--,--.
  |  .   '  ' ,-.  | | .-. | | .-. | ' ,-.  |
  |  |\   \ \ '-'  | | '-' ' | '-' ' \ '-'  |
  `--' '--'  `--`--' |  |-'  |  |-'   `--`--'
                     `--'    `--'             */





void solve() {
    set<int> digits;
    FOR (i, 0, 10)
        digits.insert(i);

    string n;
    int k;
    int g;
    cin >> n >> k;
    FOR (i, 0, k) {
        cin >> g;
        digits.erase(g);
    }

    vi v(n.size());
    FOR (i, 0, n.size()) {
        v[i] = n[i] - '0';
    }

    FOR (i, 0, v.size()) {
        auto l = digits.lower_bound(v[i]);
        if (l == digits.end()) {
            v[i] = 10;
            FOR (j, i + 1, n.size()) {
                v[j] = *digits.begin();
            }
            break;
        } else {
            if (v[i] != *l) {
                FOR (j, i + 1, n.size()) {
                    v[j] = *digits.begin();
                }
                v[i] = *l;
                break;
            }

        }
    }

    FOR (i, 0, n.size()) {
        cout << v[i];
    }


}

int main() {
    sync;


    solve();
    cerr << "Time elapsed: " << clock() / (double) CLOCKS_PER_SEC << endl;
    return 0;
}