Fill in the contents of the given memory portion at the end of the following program MOV BX, 26 MOV CX, 7 Next:MOV AL, [BX] PUSH AX INC BX LOOP Next MOV CX, 7 MOV BX, 26 Next2:POP AX MOV [BX], AL INC BX LOOP Next2 Solution ORG 00H MOV BX,26 ;initialises base register with indexed address MOV CX,7 ;initialises count register with count 7 L1:MOV AL,[BX] ;moving base register to the accumulator PUSH AX ;pushing the values into primary accumulator INC BX ;incrementing base register LOOP L1 ;it checks with count register,that if it is zero or not and go to loop l1 MOV CX,7 ;initialise count register with count 7 MOV BX,26 ;intialise base register with indexed address L2:POP AX ; poping the values from primary accumulator MOV [BX],AL ;moving al register to bx register INC BX ;increment bx LOOP L2 ;check the count register and go to loop l2 L3:PUSH AH ;pushing values into accumulator MOV AH,07H ;moving values into accumulator MOV [BX],AH ;moving values from ah register to bx register INC BH ;increment bh register DEC AH ;decrement values in ah register CMP AH,00H ;compare if ah is 0 JNE L3 ;jump to loop le if ah not equal to 0 END .