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(20160)+2(20161)+⋯+22016(20162016)(20160)+(20161)+⋯+(20162016) \dfrac{\binom{2016}{0}+2\binom{2016}{1}+\cdots+2^{2016}\binom{2016}{2016}}{\binom{2016}{0}+\binom{2016}{1}+\cdots+\binom{2016}{2016}}(02016)+(12016)+⋯+(20162016)(02016)+2(12016)+⋯+22016(20162016)
If the above expression is in the form A2016A^{2016}A2016, find AAA.
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