The following LC-3 instructions execute starting from the point shown by the comment.

1100 0000 1111 1111 ; this is the content of address D=xC0FE

1111 1010 1101 1110

; start LC-3 execution here

1010 001 111111101

0101 010 010 1 00000

0001 010 010 1 10000

0001 001 001 0 00 010

0011 001 111111111

; end LC-3 execution here

After the code reaches the end of the code (the last comment), what bits are held in R1? And in R2? And in memory locations in xC103 and xC104? If you cannot know the bits held, explain why.