Consider a language where assignments can appear in the same context as expressions; the value of a $=$ b $=$ c equals the value of c$.$ The following grammar, G$,$ generates such expressions that includes assignments in addition to additions and multiplications:

$2$

$3$

$0.$ program

$1.$ exp

$2.$ exp

$3.$ term tail

$4.$ term tail

$5.$ term

$6.$ factor tail

$7.$ factor tail

$8.$ factor

$9.$ factor

$->$ exp $$

$->$ id$=$exp

$->$ term term tail $->+$ term term tail $->\backslash$epsi

$->$ factor factor tail $->*$ factor factor tail $->\backslash$epsi

$->($exp$)$

$->$ id

$($a$)$ Showaparsetreeforthestring: id $=$ id $*($id $=$ id $+$ id $*$ id$)$$$$.$

$($b$)$ For each production A $->\backslash$alpha $,$ compute first$(\backslash$alpha $)$ and follow$($A$)$ using the algorithm below; first$(\backslash$alpha $)$ is computed by string FIRST$(\backslash$alpha $).$ For each token added, indicate the pair $($step$,$ prod$)$ used toaddit,where$0<=$step$<=3$isthestepinthealgorithm$($markedas $0,1,2,3$ below$)$and $0<=$ prod $<=9$ is

$($c$)(5$pt$)$ For each production i$,1<=$ i $<=9,$ compute predict$($i$).$

$($d$)$ Using the information computed above, show that this grammar is not LL$(1)$

$($e$)$ Modify this grammar to make it LL$(1).$ Explain clearly your changes and prove it is LL$(1).$