AWK: Printing lines after they were already processed

I have a simple file, which is processed by my awk script:

AAAAA BBBBBB
CCCCC DDDDDD
EEEEE FFFFFF
GGGGG HHHHHH
IIIII JJJJJJ

First, I want to print out one of the last lines, so I start with

#!/usr/bin/awk
{if (NR==5){print $0;}}

to get IIIII JJJJJJ as a result.

Now I want to print line 2 after line 5 was printed out. So I change my script to this:

#!/usr/bin/awk
{if (NR==5){print $0; NR=2;}
if (NR==2){print $0;}

But in this case, it is a never ending loop. How can I print line 5 first and then line 2 without printing line 5 again?

Expected output for clarity:

IIIII JJJJJJ
CCCCC DDDDDD

Answer

How about this:

awk '{if (NR==2){x=$0} else if (NR==5) {y=$0}} END {print y; print x}' input 

As we scan the file line-by-line, we store the lines we are interested in in some variables, and then we print them at the end.

And here's a slight simpler alternative:

awk '{if (NR==2){x=$0} else if (NR==5) {print $0; print x}}' input 

This only stores line 2 in a variable. When we reach line 5, we print it and then we print line 2 from the variable.

Edit: As William Pursell pointed out, a more idiomatic way to write the above code would be

awk 'NR==2 {x=$0} NR==5 {y=$0} END {print y; print x}' input 

and

awk 'NR==2 {x=$0} NR==5 {print $0; print x}' input