Perl. Format of row with number

[Nykeriab]

There are lines.

4455232
8798798324893274
74562846248376

They must be seen:

4 455 232
8 798 798 324 893 274
74 562 846 248 376

i.e. at the end of the line every time 3 The symbol insert a gap.
I can't make it.

Not the most beautiful option:

my $s = "23456779732984729374928748";
my $m = length($s)%3 ? 3 - length($s)%3 : 0 ;
my $ss = join(' ', ((' 'x$m . $s) =~ m/.../g));
$ss =~ s/^\s+//g;
print "$ss\n";

Correct option (thank you. https://ru.stackoverflow.com/users/192530/andy-37

my $s = '1923456723456779732984729374928748';
$s =~ s/(\d)(?=(\d{3})+$)/$1 /g;
print "$s\n";

[shizuka]

my $s = "12345678";
my $m = length($s)%3 ? 3 - length($s)%3 : 0 ;
my $ss = join(' ', ((' 'x$m . $s) =~ m/.../g));
print "$ss\n";

It works, but adds a gap at the beginning of 0, 1 or 2 (equalifying the length of the original line to 3N symbols). Leading gaps are easy to clean.

P.S. No demons, no disc format)

2nd method:

http://www.perlmonks.org/?node_id=110137 ♪ Totally the same, only twice used reverseto separate three symbols from the end, not from the beginning. The most elegant way from there:

my $s = '1234567';

$s =~ s/(\d)(?=(\d{3})+$)/$1 /g;

I think this way should be in the book.