Bash: Get The Last Argument Passed to a Shell Script

I'm writing a bash wrapper script that will pass arguments to the command. I need to find out the last argument if I call the wrapper as follows:

./wrapper -a -b --longarg=foo thisfilename.txt ./wrapper -a -b thisfilename.txt ./wrapper -a --next=true thisfilename.txt

Where,
=> $@ is all of them.
=> $0 is script name.
=> $1 is first arg.


I want thisfilename.txt stored in a shell variable called $last. How do I find the last argument passed to a shell script written in bash or ksh under Unix like operating systems?

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