Below is C code for a fibonacci sequence and it translated into assembly. Can you explain line by line how each was translated from C to assembly? Thanks

C code:

long fibonacci$\_$gd$\_$goto$($long $n)$

$\{$

long $i=2$;

long next, first $=0,$ second $=1$;

if $(n1)$

goto done

loop:

next $=$ first $+$ second;

first $=$ second; second $=$ next;

i$++$;

if $(in)$

goto loop;

done:

return $n$;

$\}$

Assembly Code:

fibonacci:

pushq $\%$rbp

movq $\%$rsp$,\%$rbp

movq $\%$rdi, $-40(\%$rbp$)$

movq $$2,-8(\%$rbp$)$

movq $$0,-16(\%$rbp$)$

movq $$1,-24(\%$rbp$)$

cmpq $$1,-40(\%$rbp$)$

jle $.$L$5$

nop

$.$L$2$:

movq $-16(\%$rbp$),\%$rdx

movq $-24(\%$rbp$),\%$rax

addq $\%$rdx$,\%$rax

movq $\%$rax, $-32(\%$rbp$)$

movq $-24(\%$rbp$),\%$rax

movq $\%$rax, $-16(\%$rbp$)$

movq $-32(\%$rbp$),\%$rax

movq $\%$rax, $-24(\%$rbp$)$

addq $$1,-8(\%$rbp$)$

movq $-8(\%$rbp$),\%$rax

cmpq $-40(\%$rbp$),\%$rax

jg $.$L$6$

jmp $.$L$2$

$.$L$5$:

nop

jmp $.$L$3$

$.$L$6$:

nop

$.$L$3$:

movq $-40(\%$rbp$),\%$rax

popq $\%$rbp

ret