Divisor count function

Return the number of positive divisors of n (the divisor-count function tau(n)), for n up to about 10^12. Counting by looping to n is far too slow; instead factorize n by trial division up to sqrt(n), and use that if n = product of p_i^e_i, then tau(n) = product of (e_i + 1). Return 0 for n <= 0.