$4.$ Write a program called p$3.$asm that is similar to programs $1$ and $2,$ except that this program uses IDIV to divide the first number by the second number. IDIV is similar to IMUL. However, the quotient will be stored in AL$,$ and the remainder will be stored in AH$.$ Your program should print "The quotient is: $"$ followed by the quotient on a single line, and "The remainder is: $"$ followed by the remainder on a single line. Here is the expected output.

The Dividing Program

Please enter a single digit number: $6$

Please enter a single digit number: $2$

The quotient is: $3$

The remainder is: $0$

Here is what I have done so far, I$\text{'}$m trying to divide the numbers but it won't work

BITS $32$

SECTION $.$data

startmsg db "The Dividing Program"

len$1$ equ $ $-$ startmsg

add$1$msg db "Please enter a single digit number: $",0$

len$2$ equ $ $-$ add$1$msg

add$2$msg db "Please enter a single digit number: $",0$

len$3$ equ $ $-$ add$2$msg

resultmsg db "The quotient is: $",0$

len$4$ equ $ $-$ resultmsg

remmsg db "The remainder is: $",0$

len$5$ equ $ $-$ remmsg

newline db $0$xA

SECTION $.$bss

numstore$1$ resb $1$

numstore$2$ resb $1$

result resb $1$

rem resb $1$

SECTION $.$text

global $\_$start

$\_$start:

;Starting message

mov eax, $4$ ;sys call$($sys$\_$write$)$

mov ebx, $1$ ; sys call$($std$\_$out$)$

mov ecx, startmsg ;The message

mov edx, len$1$ ; length of message

int $0$x$80$ ; call kernel

;Newline command

mov eax, $4$

mov ebx, $1$

mov ecx, newline

mov edx, $1$

int $0$x$80$

;Step $1$: Prompt the user with a single digit number

mov eax, $4$ ;sys call$($sys$\_$write$)$

mov ebx, $1$ ; sys call$($std$\_$out$)$

mov ecx, add$1$msg ; The message

mov edx, len$2$ ; length of message

int $0$x$80$ ; call kernel

;Read in the first number

mov eax, $3$

mov ebx, $0$

mov ecx, numstore$1$

mov edx, $2$

int $0$x$80$

;Prompt user for the second number

mov eax, $4$

mov ebx, $1$

mov ecx, add$2$msg

mov edx, len$3$

int $0$x$80$

;Read in the second number

mov eax, $3$

mov ebx, $0$

mov ecx, numstore$2$

mov edx, $2$

int $0$x$80$

;Convert the first number from ASCII to integer

mov ax$,[$numstore$1]$

sub ax$,\text{'}0\text{'}$

;Convert the second number from ASCII to integer then add to first number

mov bx$,[$numstore$2]$

sub bx$,\text{'}0\text{'}$

;Divide the second number from the first number

idiv ax

;Convert the answer from a number to a string

add al$,\text{'}0\text{'}$

add ah$,\text{'}0\text{'}$

mov $[$result$],$ al

mov $[$rem$],$ ah

;Print result message

mov eax, $4$

mov ebx, $1$

mov ecx, resultmsg

mov edx, len$4$

int $0$x$80$

;Print the result

mov eax, $4$

mov ebx, $1$

mov ecx, result

mov edx, $2$

int $0$x$80$

;Print remainder message

mov eax, $4$

mov ebx, $1$

mov ecx, remmsg

mov edx, len$5$

int $0$x$80$

;Print the remainder

mov eax, $4$

mov ebx, $1$

mov ecx, rem

mov edx, $1$

int $0$x$80$

mov eax, $4$

mov ebx, $1$

mov ecx, newline

mov edx, $1$

int $0$x$80$

mov eax, $1$

int $0$x$80$ properly.