{   My program
task
7
2 3 0
3 1 0
1 2 4 5 0
3 0
3 0
7 0
6 0
answer
4
1 3 5 7
it is not the same as standart answer but it is also right
}

{$APPTYPE CONSOLE}
var
   fr:array[1..100,0..99] of byte;
   A,B,mem:set of byte;
   n,i,j,kol:integer;
   group:char;
label bed;
procedure getfriend(n:integer);
var i:integer;
begin
  if (n in mem) and (not (n in a)) and (not (n in b)) then
 begin
  mem:=mem-[n];
  if group='A'
  then
  begin
     inc(kol);
     a:=a+[n];
     group:='B'
  end
  else
  begin
    b:=b+[n];
    group:='A';
  end;
  for i:=1 to fr[n,0] do
  getfriend(fr[n,i]);
 end
end;
begin
 { assign(Input,'input.txt'); assign(Output,'output.txt');
  reset(Input);              rewrite(Output);             }
  group:='A';
  readln(n);
  for I:=1 to n do
  begin
    j:=0;
    while not eoln do
    begin
      inc(j);
      read(fr[i,j]);
    end;
    if j=1 then
    begin
       writeln(0);
       goto bed;
    end;
    readln;
    fr[i,0]:=j-1;
  end;
  mem:=[1..n];
  for i:=1 to n do
  getfriend(i);
  writeln(kol);
  for i:=1 to n do if i in A then write(I,' ');
  writeln;
  bed:
{  close(Input);              close(Output);}
end.

The 5 he has the friend 3, and with him in one command. In an opposite command the 5 does not have friends. Your answer false!!!