For a positive integer $n$, let $f(n)$ be the number of (not necessarily distinct) primes in the prime factorization of $k$. For example, $f(1) = 0$, $f(2) = 1$, and $f(4) = f(6) = 2$. Let $g(n)$ be the number of positive integers $k \leq n$ such that $f(k) \geq f(j)$ for all $j \leq n$. Find $g(1) + g(2) + \ldots + g(100)$.

Let's think step by step and output the final answer within \boxed{}.
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