Let an ordered pair of positive integers $(m, n)$ be called regimented if for all nonnegative integers $k$, the numbers $m^k$ and $n^k$ have the same number of positive integer divisors. Let $N$ be the smallest positive integer such that $(2016^{2016}, N)$ is regimented. Compute the largest positive integer $v$ such that $2^v$ divides the difference $2016^{2016} - N$.

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