Get date from cal function

Cal
Sun Mon Tue Wed Thu Fri Sat
1 2 3 4 5 6
7 8 9 10 11 12 13
14 15 16 17 18 19 20
21 22 23 24 25 26 27
28 29 30

cal | awk '{print $7}'
Sat

13
20
27
In the above output i am missing 6 because the first column is empty so it is shfting.
How can i print all the dates under sat.
I need the output like below.
Sat
6
13
20
27

Use cut to cut out the column by character count.

Use nawk or /usr/xpg4/bin/awk on Solaris:

cal|awk '{print/^  /?$NF:$7}'

A very ugly way to do this...But the solution suggested by radoulov must be the best way..

cal | sed -n '/\(.*\) \(.*\) \(.*\) \(.*\) \(.*\) \(.*\) \(.*\)/p' |sed 's/\(.*\) \(.*\)/\2/'