sed to output blocks , which contain a pattern

Hi Friends,

I need a help in using the sed .

Objective is to search and output the block which contains a pattern 
inside 
the block.

I have a block eg.

BEGIN
aaaa
bbbb
ccc
ddd
END
BEGIN
aaaa
bbbb
ddd
END
BEGIN
aaaa
bbbb
ddd
END

I need to output the block ( from BEGIN to END ) which contains the 
pattern 
"ccc" in it , hence I need the out put to be


BEGIN
aaaa
bbbb
ccc
ddd
END

Could any one suggest how to go about this

Thanks..Siva






------------------------ Yahoo! Groups Sponsor --------------------~--> 
Most low income homes are not online. Make a difference this holiday season!
http://us.click.yahoo.com/5UeCyC/BWHMAA/TtwFAA/dkFolB/TM
--------------------------------------------------------------------~-> 

-- 
 
Yahoo! Groups Links

<*> To visit your group on the web, go to:
    http://groups.yahoo.com/group/sed-users/

<*> To unsubscribe from this group, send an email to:
    sed-users-unsubscribe@xxxxxxxxxxxxxxx

<*> Your use of Yahoo! Groups is subject to:
    http://docs.yahoo.com/info/terms/