What does the followiing program do$?$

pop eax; get the return address

add eax, $3$ ; add $3$

push eax; put it back on the stack

ret

return to an address that was $3$ bytes higher in memory than the return address currently on the stack

return to an address that was $3$ bytes lower in memory than the return address currently on the stack

return to an address that was $6$ bytes higher in memory than the return address currently on the stack

return to an address that was $12$ bytes higher in memory than the return address currently on the stack