For the next three questions, refer to this assembly program, and assume little endian. Note that each instruction takes $4-$bytes.

Address $|$ Code

$0$x$00000000|$ main: MOVS R$0,$ #$0$x$3$

$0$x$00000004|$ BEQ end

$0$x$00000008|$ loop: BL fun$1$

$0$x$0000000$C $|$ CMP R$0,$ #$0$x$0$

$0$x$00000010|$ BGT loop

$0$x$00000014|$ end: B end

$0$x$00000018|$ fun$1$: SUB r$0,$ #$0$x$1$

$0$x$0000001$C $|$ PUSH $\{$LR$\}$

$0$x$00000020|$ BL fun$2$

$0$x$00000024|$ POP $\{$LR$\}$

$0$x$00000028|$ BX LR

$0$x$0000002$C $|$ fun$2$: MOV R$8,$ #$0$x$0000$

$0$x$00000030|$ MOVT R$8,$ #$0$x$2000$

$0$x$00000034|$ LSL R$1,$ R$0,$ #$2$

$0$x$00000038|$ ADD R$8,$ R$1$

$0$x$0000003$C $|$ STR R$0,[$R$8]$

$0$x$00000040|$ BX LR

Note: LSL means "Logical Shift Left." The line of code at address $0$x$34$ is taking the value in R$0,$ shifting the bits to the left $2$ times, then placing the result in R$1.$ Shifting to the left n times is the same as multiplying by $2^$n$.$

$($For this question, use the same code above$)$

Assume PC is currently set to: $0$x$00000008$

On the next time step $($in other words, after the instruction at PC $=0$x$00000008$ executes$),$ what will the contents of the following registers be$?$

enter all $32$ bits for each, written in hexadecimal with $0$x included.

PC:

LR: