can someone please help me fix my MIPS CODE so that it will print the results as shown in the picture. This is the code:

# Project $3$ Part B Template

# Cube root of floating point numbers in Single Precision IEEE floating point format

$.$data

zeroFloat: $.$float $0\mathrm{.}0$

threefloat: $.$float $3\mathrm{.}0$

hdeFloat: $.$float $8\mathrm{.}0$ # Hardcoded half debiased estimate of user input s$0($obtained from Part$-$A$)$

newLine: $.$asciiz $"$

$"$

$.$text

# For output purposes

lwc$1$ $f$10,$ zeroFloat # $f$10=0\mathrm{.}0$

lwc$1$ $f$8,$ hdeFloat # $f$8=8\mathrm{.}0$

# read float syscall, the float to find the cube root of

li $v$0,6$ # read float

syscall

mtc$1$ $t$0,$ $f$0$ # copy Co$-$proc $1$ reg. $f$0$ to reg. $t$0|$ $t$0=$ input number

mtc$1$ $s$0,$ $f$0$ # copy Co$-$proc $1$ reg. $f$0$ to reg. $s$0|$ $s$0=$ input number $|$ to calculate half debiased estimate $($HDE$)$

# the HDE value serves as the first educated guess for the Newton's Method approximation

mfc$1$ $t$2,$ $f$8$

# read integer syscall, the number of iterations to run Newton's method

li $v$06$ # read integer

syscall

move $t$1,$ $v$0$ # move the integer read into $t$1|$ $t$1=$ iterations

# loop for calling Newton's method $t$1$ times

loop:

jal cbrt # call cbrt to calculate the next iteration of Newton's Method

mtc$1$ $t$2,$ $f$4$ # copy answer of Newton's estimate from $t$2$ to $f$4$

# print the output of the current iteration of Newton's Method as a float

li $v$0,2$ # print float

add.s $f$12,$ $f$4,$ $f$10$ # $f$12=$ $f$4+$ $f$10|$ $f$12=$ $f$4$

syscall

# print a new line

li $v$0,4$ # print string

la $a$0,$ newLine # load new line string

syscall

# decrement iteration counter $t$1$

subi $t$1,$ $t$1,1$ # $t$1=$ $t$1-1$

beq $t$1,$ $$0,$ exit # if $t$1==0,$ exit the program

bne $t$1,$ $$0,$ loop # else loop again

exit:

# exit the program

li $v$0,10$ # exit

la $a$0,($$v$0)$ # load exit

syscall

cbrt:

mtc$1$ $t$0,$ $f$0$ # copy original user input number to find cube root of from $t$0$ to Co$-$proc $1$ reg $f$0$

mtc$1$ $t$2,$ $f$1$ # copy the educated guess approximation HDE value from $t$2$ to Co$-$proc $1$ reg. $f$1$

lwc$1$ $f$2,$ threefloat # load $3\mathrm{.}0$ into Co$-$proc $1$ reg. $f$2$

# x$\_$n$+1$ x$\_$n $-($x$\_$n$^3-$ N$)/(3*$ x$\_$n$^2)$

NewtonsMethod:

mul.s $f$5,$$f$1,$$f$1$ # $f$5=$ $f$1*$ $f$1,$ i$.$e$.$ $f$5=$ x$^2$

mul.s $f$5,$ $f$5,$$f$1$ # $f$5=$ $f$5*$ $f$1,$ i$.$e$.$ $f$5=$ x$^3$

sub.s $f$6,$ $f$5,$ $f$0$ # $f$6=$ $f$5-$ $f$0$

div.s $f$7,$ $f$5,$ $f$2$ # $f$7=($x$^3-$ N$)/3$

mul.s $f$4,$ $f$1,$ $f$1$ # $f$4=?$

div.s $f$8,$ $f$7,$$f$4$ # $f$4=($x$^3-$ N$)/(3*$ x$^2)$

add.s $f$4,$ $f$1,$ $f$7$ # $f$4=$ x$\_$n $-[($x$\_$n$^3-$ N$)/(3*$ x$\_$n$^2)]$

# copy answer for the next iteration of Newton's Method from Co$-$proc $1$ reg. $f$4$ to reg. $t$2$

mfc$1$ $t$2,$ $f$4$ # $t$2=$ answer

# Netwon's Method sqrt estimate answer is in $f$4$

jr $ra