Trying to answer this in C++

uestion 5: Consider the following definitions a. A proper divisors of a positive integer (2 2) is any of its divisors excluding the number itself For example, the proper divisors of 10 are: 1, 2 and 5 A perfect number is a positive integer (2 2) that is equal to the sum of its proper divisors For example, 6 and 28 are perfect numbers, since 6=1+2+3 28 1+2+4+7+14 Background of perfect numbers: https://en.wikipedia.org/wiki/Perfect number b. c. Amicable numbers are two different positive integer (2 2), so related that the sum of the proper divisors of each is equal to the other number For example, 220 and 284 are amicable numbers, since 284 = 1 + 2 + 4 + 5 10 + 11 + 20 + 22 44 55 + 110 220 1 + 2 + 4 Z1+ 142 (proper divisors of 220) (proper divisors of 284) Background of amicable numbers: https://en.wikipedia.org/wiki/Amicable_numbers Write a function void analyzeDividors (int num, int& outCountDivs, int& outSumDivs) The function takes as an input a positive integer num (2 2), and updates two output parameters with the number of num's proper divisors and their sum For example, if this function is called with num=12, since 1, 2, 3, 4 and 6 are 12s proper divisors, the function would update the output parameters with the numbers 5 and 16 Note: Pay attention to the running time of vour function. An efficient implementation would run in (Wtum) a. b. Use the function you wrote in section (a), to implement the function bool isPerfect (int num) This functions is given positive integer num (2 2), and determines if it is perfect number or not. Use the functions you implemented in sections (a) and (b), to write a program that reads from the user a positive integer M (2 2), and prints c. All the perfect numbers between 2 and M All pairs of amicable numbers that are between 2 and M (both numbers must be in the range) Note: Pay attention to the running time of your implementation. An efficient algorithm for this part would call analyzeDividors (M) times all together