i want to know how to reverse first 3 bits of AX IN this code to get the anew character from memory in emu 8086? please quickly help

include 'emu8086.inc'

#BX=0000h# #DI=0000h# #DS=0000h#

MOV CX,7d

AGAIN:NOP

MOV AH,01H

INT 21H MOV [DI],Al

MOV AX,DI INC DI

CALL PRINT_NUM

DEFINE_PRINT_NUM DEFINE_PRINT_NUM_UNS LOOP AGAIN MOV SI,0f8H

MOV CX,7d GOTOXY 1,2

D1:NOP MOV AX,SI

INC SI CALL PRINT_NUM

mov DI,AX

mov ah,2h mov DL,[DI] INT 21H

LOOP D1