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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var
    n,i,j,m,k,d,t,t0,bit_count,bit_need,s,first_ava: Integer;
    flag,flag2:Boolean;
    bit:array[1..4] of Integer;
    bit_avaliable:array[0..9] of Boolean;
    match:array[1..6] of Integer;
    match2:array[1..6] of Integer;
    bit_out:array[1..5] of Integer;

procedure my_output(n:Integer);
var
    i: Integer;
begin
    for i := 1 to n do
    begin
        write(bit_out[i]);
    end;
    writeln;
    readln;
    readln;
end;

begin
    readln(n,k);
    for i := 0 to 9 do
    begin
        bit_avaliable[i]:=true;
    end;
    for i := 1 to k do
    begin
        read(d);
        bit_avaliable[d]:=false;
    end;

    bit_count:=0;
    while n>0 do
    begin
        inc(bit_count);
        bit[bit_count]:=n mod 10;
        n:=n div 10;
    end;
    bit_need:=bit_count;

    t:=0;
    for j := bit_count downto 1 do
    begin
        flag2:=true;
        for i := bit[j] to 9 do
        begin
            if bit_avaliable[i] then
            begin
                flag2:=false;
                inc(t);
                match[t]:=i;
                flag:=true;
                for m := i+1 to 9 do
                begin
                    if bit_avaliable[m] then
                    begin
                        match2[t]:=m;
                        flag:=false;
                        break;
                    end;
                end;
                if flag then
                begin
                    match2[t]:=-1;
                end;
                break;
            end;
        end;
        if flag2 then
        begin
            t0:=t;
            while (t0>0) and (match2[t0]=-1) do
            begin
                dec(t0);
            end;
            if t0=0 then
            begin
                inc(bit_need);
                for i := 1 to 9 do
                begin//get first avaliable
                    if bit_avaliable[i] then
                    begin
                        first_ava:=i;
                        break;
                    end;
                end;
                for s := 1 to bit_need do
                begin
                    bit_out[s]:=first_ava;
                end;
                my_output(bit_need);
                exit;
            end
            else
            begin
                for s := 1 to bit_need do
                begin
                    bit_out[s]:=match[s];
                end;
                bit_out[t0]:=match2[t0];
                my_output(bit_need);
                exit;
            end;
        end;
    end;

    for s := 1 to bit_need do
    begin
        bit_out[s]:=match[s];
    end;
    my_output(bit_need);
end.