perl、shell格式化文本（70字符每行转6列每列10字符）

如果我想对一个每行70个字符的文本处理，让它生成每列10个字符，每行6列，并输出字符数怎么弄呀？
如：

1. TINHTFGNPVIPFKDGIYFAATEKSNVVRGWVFGSTMNNKSQSVIIINNSTNVVIRACNFELCDNPFFAV
2. SKPMGTQTHTMIFDNAFNCTFEYISDAFSLDVSEKSGNFKHLREFVFKNKDGFLYVYKGYQPIDVVRDLP

生成：

1. 1        TINHTFGNPV        IPFKDGIYFA        ATEKSNVVRG        WVFGSTMNNK        SQSVIIINNS       TNVVIRACNF
2. 61        ELCDNPFFAV        SKPMGTQTHT        MIFDNAFNCT        FEYISDAFSL        DVSEKSGNFK       HLREFVFKNK
3. 121        DGFLYVYKGY        QPIDVVRDLP

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fold -w10 urfile |xargs -n6 |awk '{print (NR-1)*60+1,$0}'

#!/usr/bin/perl -w

my ( $column, $character ) = ( 5, 15 );    # ( 6, 10 )
my ( $num, $X, $data ) = ( 0, $column * $character );
while () { chomp; $data .= $_ }
my @split = $data =~ /(.{$character}|.+)/g;

while ( my @line = splice @split, 0, $column ) {
    print join "\t", $num++ * $X + 1, @line;
    print "\n";
}
——————————————————————————————————————————
my $i = 0;
while () {
    my @line = unpack "A10" x ( length($_) );
    my @data;
    for (@line) {
        next unless $_;
        print 1, "\t" if $i == 0;
        $i += length($_);
        print $_, "\t";
        print "\n", $i + 1, "\t" if $i % 60 == 0;
    }
}