






import math
def      lowerCamelCase					(						a_	)	->			Dict:
							lowerCAmelCase_										=     0
							if n % 2 == 0:
														count += 1
														while n % 2 == 0:
																					lowerCAmelCase_										=     int(n / 2	)
    # the n input value must be odd so that
    # we can skip one element (ie i += 2)

							lowerCAmelCase_										=     3

							while i <= int(math.sqrt(a_	)	):
														if n % i == 0:
																					count += 1
																					while n % i == 0:
																												lowerCAmelCase_										=     int(n / i	)
														lowerCAmelCase_										=     i + 2

							# this condition checks the prime
							# number n is greater than 2

							if n > 2:
														count += 1
							return count


if __name__ == "__main__":
		lowerCamelCase_						     =  5_1_2_4_2_1_8_3
		print(f'''The number of distinct prime factors is/are {exact_prime_factor_count(n)}''')
		print(f'''The value of log(log(n)) is {math.log(math.log(n)):.4f}''')











from doctest import testmod
from math import sqrt
def      lowerCamelCase					(						a_	)	->			list:
							lowerCAmelCase_										=     []
							if num < 1:
														return facs
							facs.append(1	)
							if num == 1:
														return facs
							facs.append(a_	)
							for i in range(2 ,						int(sqrt(a_	)	) + 1	):
														if num % i == 0:  # If i is a factor of num
																					facs.append(a_	)
																					lowerCAmelCase_										=     num // i  # num//i is the other factor of num
																					if d != i:  # If d and i are distinct
																												facs.append(a_	)  # we have found another factor
							facs.sort()
							return facs


if __name__ == "__main__":
		testmod(name="""factors_of_a_number""", verbose=True)



