Write a EMU 8086 program to add two 8-digit BCD numbers stored in AX:BX and CD:DX registers. The sum should be placed in CX:DX registers. Your solution should be generic and should product correct results even if the values in registers are changed.

void main()
{
  // BCD 0000 0000 0000 0000
  const short m = 0x0000;
  const short n = 0x0000;
  
  short carry = 0x0000;
  short a = m & 0x000F;
  short b = n & 0x000F;
  short a1 = (m >> 4) & 0x000F;
  short b1 = (n >> 4) & 0x000F;
  short a2 = (m >> 8) & 0x000F;
  short b2 = (n >> 8) & 0x000F;
  short a3 = (m >> 12) & 0x000F;
  short b3 = (n >> 12) & 0x000F;
  
  short p = 0;
  short q = 0;
  short r = 0;
  short s = 0;
  
  p = a + b + carry;
  if (p > 9) {
    carry = 1;
    p = p - 10;
  } else {
    carry = 0;