2.29 [5] Translate the following loop into C. Assume that the C-level integer i is held in register $t1,$s2 holds the C-level integer called result, and $50 holds the base address of the integer MemArray. addi $t1, $ 0, $0 LOOP: lw $51, 0($0) add $52, $52, $51 addi $50, $50, 4 addi $t1, $t1, 1 slti $t2, $t1, 100 bne $t2, $50, LOOP