Consider this MIPS code:

loop:lw $8, 0($16)

lw $9, 4($16)

add $8, $9, $8

sw $8, 0($17)

addi $16, $16, 8

addi $17, $17, 4

bne $16, $4, loop

Unroll 3 times the MIPS code (i.e., one unrolled iteration does the work of four original iterations). Make it as efficient as you can. Show timing in the 5-stage MIPS pipeline for one unrolled iteration of the loop. Assume that the number of iterations is always a multiple of four