Given a positive integer $n$ with prime factorization $p_1^{e_1}p_2^{e_2}... p_k^{e_k}$ , we define $f(n)$ to be  $\sum^k_{i=1}p_ie_i$. In other words, $f(n)$ is the sum of the prime divisors of $n$, counted with multiplicities. Let $M$ be the largest odd integer such that $f(M) = 2023$, and $m$ the smallest odd integer so that $f(m) = 2023$. Suppose that $\frac{M}{m}$ equals $p_1^{e_1}p_2^{e_2}... p_l^{e_l}$ , where the $e_i$ are all nonzero integers and the $p_i$ are primes. Find $\left| \sum^l_{i=1} (p_i + e_i) \right|$.

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