Building scalar from File::Find

sunadmn has asked for the wisdom of the Perl Monks concerning the following question:

Good day fellow monks I have the below code and have a it working for the most part but one issue I have ran into is when I build $out if there is not seven dirs I get output that looks like so:

net.somedomain......

Now the issue I would like to resolve is how can I build my $out scalar so that it will grab anything from dir 4 on and build itself from that,so if I have a file living under /net it will build $out like somedomain.net and if I have something like /us/pa/k12 it will look like somadomain.k12.pa.us and with either I wont have the trailing "." after the name if it doesn't equal seven in length. Any help you may all point out to me would be great as I am pulling my hair out trying to figure this one out right now.

Sunadmn
USE PERL

#!/usr/bin/perl

use strict;
use File::Find;
use FileHandle;

my $base    = "/chroot/named/master";
my $out     = "/var/tmp/current_list";



find(\&wanted, $base);

sub wanted {
   return unless ( -f "$File::Find::name" and $File::Find::name !~ m!/
+in-addr/! );

   my $name  = "$File::Find::name";
   my $zone  = new FileHandle "$name", "r";
   my $out   = join('.', ( split('/', $name) )[4..7]);
   print "$out\n";
}
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