Based on the figure below, implement two four bits adder

using port$-$ D as input for number A and number B $,$ and port$-$

A as output for result:

$>$ Show the inputs and outputs interface with the ports.

$>$ Develop the program to read the inputs, add the two

numbers then display the result on the lower four LEDs

of port A and LCD Line $1$ Sum $=$ xx$.$

$>$ If the answer greater than $0$ XOF $,$ display the number as

before, in addition to set the Cout $($port A bit $4)$ and on

Line $2$ of the LCD display "Carry Flag Set""$].$

I am not able to upload an image but its a $4-$bit adder

pls explain step by step i need the program like this:

#include

#include"config.h$"$

#include "LCD.h$"$

int main$($void$)$

$\{$

TRISA$=0$X$0000$;

TRISDbits.TRISD$7=1$;

TRISAbits.TRISA$7=1$;

TRISDbits.TRISD$13=1$;

TRISDbits.TRISD$6=1$;

int num$1,$ num$2,$ sum;

initLCD$()$;

while$(1)$

$\{$

homeLCD$()$;

num$1=$ PORTDbits.RD$6*1+$ PORTDbits.RD$7*2$;

num$2=$ PORTAbits.RA$7*1+$ PORTDbits.RD$13*2$;

sum $=$ num$1+$num$2$;

PORTA $=$ sum;

$14$

setLCDC$($LINE$1)$;

putsLCD$("$Sum $=")$;

putdLCD$($sum$)$;

setLCDC$($LINE$2)$;

if$($sum$>3)$

$\{$ PORTAbits.RA$3=1$;

putsLCD$("$Carry Set"$)$;

$\}$

else

$\{$PORTAbits$.$RA$3=0$;

putsLCD$("")$;

$\}$

$\}$

return $0$;

$\}($This is only an example of how i want it$,$ this code is a $2-$bit adder$)$