var i,n,L,R:longint;
    ch:char;
begin
 readln(n);
 L:=0;
 R:=0;
 for i:=1 to n do
 begin
  read(ch);
  while not(ch in ['<','>']) do
   read(ch);
  case ch of
   '>':inc(R);
   '<':inc(L,R);
  end;
 end;
 writeln(L);
end.

Denote by R the number of recruits at the right end of the queue. When you get a new '>', simply increase R by 1. If you receive a '<', the process is like this:

>>>>>< (Say R=5)
>>>><>
>>><>>
>><>>>
><>>>>
<>>>>>

Now you see, it takes exactly R turns to send the '<' to the left of the '>'s, and the number of the '>'s, or R, stays the same. So it's easy to understand the prog.

10
>>>><>>><

What about this test?

thanks