a. Write a line by line explanation for the following program.b. Write a short paragraph on what is the logic of the programc. What will be displayed on the screen when the program terminates? .model smally proc.386t:.stack 100hmov ah, 1.dataint 21hmsg db "W ELL DONE GOOD BYE!", 13, "$"", "$"cmp al, pppmsg1 dbYU db 2jnz tSHL PPP, 1ppp db 27.codeinc pppretmain procy endpmov ax,@datamov ds,axmain endpL:s procMOV AH,9call ydec YUINT 21Hjnz Lrets endpLEA DX, MSGcall send maincall yLEA DX, MSG1call smov ax, 4c00hint 21h

Actually, assembly language is a low low-level language. It uses mnemonic codes and labels to represent machine level code. MODEL SMALL ;use the small memory model(one sgment for stack, data and code) .STACK 100H ;reserve 100 bytes for stack .data ; data section nono db 8 ;byte nono with value 8 c db 0 ;byte c with value 0 bsd db 0 ;byte bsd with value 0 x db "Enter any key",13,10,"$" ;string x .code ;code segment main proc mov AX,@data ;intitialize data segment mov ds,ax mov dx,offset x ;move address of string x into dx call mssg ;call procedure mssg to display string x on screen call y ;call procedure y , to read key from user mov bsd,al ;al=bsd,bsd=any capital/upper case letter read mov c,0 ;c=0 mm: shr bsd,1 ;shift bits in bsd right one time,move lsb to CF ;if lsb=0,CF=0 ese CF=1 jc short jjj ;if carry(CF=1),if lsb=1, jump to label jjj inc c ;if(CF=0), if(lsb=0), increment c, c=c+1 ;c counts number of 0's in bsd jjj: dec nono ;decrement nono, nono=nono-1. sets ZF=1, when nono=0 jnz mm ;if(ZF=1), i.e. if (nono !=0) goto label mm ;shft bsd right till nono=0, i.e 8 times mov dl,c ;count of 0's in c is moved to dl shr c,1 ;shift c one time right,if lsb=1(odd number),CF=1 jc short start1 ;if(CF=1), count of 0's is odd,goto start1, to terminate call z ;if CF=0, count 0f 0's even, call z to print the count start1: mov ax,4c00h ;"EXIT" - TERMINATE WITH RETURN CODE int 21h main endp y proc ;procedure y which reads key till user pressers any ;letter between A and Z uuu: mov ah,1 ;function ah=01 used to read character int 21h ; the character read is in al cmp al,65 ;if character is A (ascii value of A=65) js uuu ;if character < 'A', then goto uuu to repeat read cmp al,91 ;compare al and 91, 91 is the ascii vlue of [ (Z=90) jns uuu ;if character >'Z', theh goto uuu to repeat read ret ;return from procedure y endp ;end of procedure y z proc ;procedure z to display the count of 0's if even mov ah,6 ;function ah=06 to display character in dl add dl,48 ; convert the number in dl to character by adding 48 to dl int 21h ret z endp mssg proc ;procedure mssg to display string mov ah,9 ;function ah=9 to display strting(its address in dx) int 21h ret ;return from procedure mssg endp ;end of procedure ;main endp end main Program prompts the user to enter any key Then it accepts the keys until the user inputs any character between A and Z (Only upper case letter) Then store the input character in al into variable bsd Shift bsd one time right and count the number of 0's in the charcater ( If lsb=0, then CF=0, increment count of 0's in c) The shift repeats for 8 times (initially nono=8, in each iteration it decrements and repeats till nono=0) Now the c has count of 0's in the input uppercase letter and move it to dl Then check the count is even (shift right c. if lsb=0(even), CF=0), then display the count by calling procedure z. The proedure z , convert the count into character by adding 48 to number. (e.g. '0' = 0 + 48) If count is odd, not displaying anything and terminate Problem Definition: Accept a uppercase letter from the user. Count the number of 0's in the character(ASCII value of character). If the count is even, display the same.