> program P1002;
>   const
>     max=100000;
>   type
>     ar=record
>          num,times:longint;
>        end;
>   var
>     same,m,n,number,times:longint;
>     call:array[1..max] of ar;
>   procedure getnumber;
>     var
>       i,k,code:integer;
>       base:longint;
>       strings,rest:string;
>     begin
>       readln(strings);
>       rest:='';
>       for i:=1 to length(strings) do
>         if (strings[i] in ['0'..'9'])or
>            (strings[i] in ['A'..'Y']) then
>           rest:=rest+strings[i];
>       for i:=1 to length(rest) do
>         if rest[i] in ['A'..'Y'] then
>           if rest[i]<'Q' then
>             rest[i]:=chr(((ord(rest[i])-65) div 3+2)+48) else
>             rest[i]:=chr(((ord(rest[i])-66) div 3+2)+48);
>       base:=1000000;
>       number:=0;
>       for i:=1 to 7 do
>         begin
>           val(rest[i],k,code);
>           number:=number+longint(base*k);
>           base:=base div 10;
>         end;
>     end;
>   function min(x,y:longint):longint;
>     begin
>       if x<y then
>         min:=x else
>         min:=y;
>     end;
>   procedure builtheap;
>     var
>       mid:ar;
>       k:longint;
>     begin
>       call[n].num:=number;
>       call[n].times:=1;
>       k:=n;
>       while (k div 2>0)and(call[k div 2].num>=call[k].num) do
>         begin
>           if call[k div 2].num>call[k].num then
>             begin
>               mid:=call[k div 2];
>               call[k div 2]:=call[k];
>               call[k]:=mid;
>               k:=k div 2;
>             end                            else
>           if call[k div 2].num=call[k].num then
>             begin
>               inc(call[k div 2].times);
>               call[k]:=call[n];
>               call[n].num:=0;
>               call[n].times:=0;
>               dec(n);
>               while (call[k].num>min(call[2*k].num,call[2*k+1].num))and
>                     (2*k<=n) do
>                 begin
>                   if (2*k+1>n)or
>                      ((2*k+1<=n)and(call[2*k].num<call[2*k+1].num)) then
>                     begin
>                       mid:=call[k];
>                       call[k]:=call[2*k];
>                       call[2*k]:=mid;
>                       k:=2*k;
>                     end      else
>                     begin
>                       mid:=call[k];
>                       call[k]:=call[2*k+1];
>                       call[2*k+1]:=mid;
>                       k:=2*k+1;
>                     end;
>                 end;
>               break;
>             end;
>         end;
>     end;
>   procedure getmin;
>     var
>       mid:ar;
>       k:longint;
>     begin
>       number:=call[1].num;
>       times:=call[1].times;
>       k:=1;
>       call[k]:=call[n];
>       call[n].num:=0;
>       call[n].times:=0;
>       dec(n);
>       while (call[k].num>min(call[2*k].num,call[2*k+1].num))and
>             (2*k<=n) do
>         begin
>           if (2*k+1>n)or
>              ((2*k+1<=n)and(call[2*k].num<call[2*k+1].num)) then
>             begin
>               mid:=call[k];
>               call[k]:=call[2*k];
>               call[2*k]:=mid;
>               k:=2*k;
>             end      else
>             begin
>               mid:=call[k];
>               call[k]:=call[2*k+1];
>               call[2*k+1]:=mid;
>               k:=2*k+1;
>             end;
>         end;
>     end;
>   procedure print;
>     var
>       i:integer;
>       base:longint;
>     begin
>       base:=1000000;
>       for i:=1 to 7 do
>         begin
>           write(number div base);
>           number:=number mod base;
>           base:=base div 10;
>           if i=3 then
>             write('-');
>         end;
>       writeln(' ',times);
>       inc(same);
>     end;
>   procedure doing;
>     var
>       i:longint;
>     begin
>       same:=0;
>       for i:=1 to max do
>         begin
>           call[i].num:=0;
>           call[i].times:=0;
>         end;
>       readln(m);
>       n:=0;
>       for i:=1 to m do
>         begin
>           getnumber;
>           inc(n);
>           builtheap;
>         end;
>       m:=n;
>       times:=0;
>       for i:=1 to m do
>         begin
>           getmin;
>           if times>1 then
>             print;
>         end;
>       if same=0 then
>         writeln('No duplicates.');
>     end;
>   begin
>     {assign(input,'1002.in');
>     reset(input);}
>     doing;
>     {close(input);}
>   end.