Consider the following delay subroutine:

delay: movlw fah

movwf delayreg

loop: nop

nop

decfsz delayreg,$1$

goto loop

return

If this was written for a PIC$16$f$88$ running from a $4$ MHz clock signal:

Calculate the delay it actually introduces. $(12)$

Modify the given delay subroutine to introduce a time delay of $10$ ms$.$