module test$($in$,$slt$,$out$)$; input in$,[1$:$0]$slt; output reg $[3$:$0]$ out; always @ $($in or slt$)$ begin case $($slt$)2\text{'}$b$00$: out$[0]=$in; $2\text{'}$b$01$: out$[1]=$in; $2\text{'}$b$10$: out$[2]=$in; $2\text{'}$b$11$: out$[3]=$in; endcase end endmodule A$.$ Give the value of the output $($value of each bit$)$ for the following input values: in$=0,$ slt$[1$:$0]=10$ in$=0,$ slt$[1$:$0]=01$ in$=1,$ slt$[1$:$0]=10$ B$.$ If we modify the always block as follows: always @ $($in and slt$)$ begin case $($slt$)2\text{'}$b$00$: out$[0]=$in; $2\text{'}$b$01$: out$[1]=$in; $2\text{'}$b$10$: out$[2]=$in; $2\text{'}$b$11$: out$[3]=$in; endcase end What will be the values of the output for the following input values? in$=0,$ slt$[1$:$0]=10$ in$=0,$ slt$[1$:$0]=01$ in$=1,$ slt$[1$:$0]=10$ C$.$ The code given in part a$)$ corresponds to a de$-$multiplexer where one output is selected by the selector to get the value of the input. If we add one more bit for the slt input, in a way that the chosen output corresponds to the modulus of the slt input value, ex$.$: if the slt input value is $101($i$.$e$.5$ in decimal$),$ the selected output will be out$[3]$ because $3\%8=3.$ Mo