A MIPS assembly code is shown below. The assembly code is executed using a $5-$stage MIPS processor and we can actually decide the branch a little earlier, in ID instead of EX$.$ Answer the following questions: Loop: sub $$19,$ $$19,$ $$20$ addi $$9,$ $$20,5$ sw $$9,0($$$21)$ addi $$10,$ $$19,2$ sll $$10,$ $$10,$ $$3$ add $$10,$ $$10,$ $$21$ addi $$11,$ $$20,1$ sll $$11,$ $$11,3$ add $$11,$ $$11,$ $$21$ lw $$11,0($$$11)$ sub $$9,$ $$9,$ $$11$ sw $$9,0($$$10)$ beq $$9,$ $$0,$ Loop addi $$9,$ $$12,2$a$).$ How many clock cycles are needed to execute the code above? Assume there is no forwarding, no branch prediction.b$).$ How many clock cycles are needed to execute the code above? Assume there is forwarding, no branch prediction.c$).$ Reorder the above code to reduce the stalls $($get the smallest number of stalls$)$ without changing the functionality. Assume there is forwarding and no branch prediction. How many clock cycles are needed to execute the code above?