Let $p(x)$ be a polynomial of degree 100 such that
\begin{align*}
p(1) &= 1, \\
p(2) &= 2, \\
p(3) &= 3, \\
&\dots, \\
p(99) &= 99, \\
p(100) &= 100, \\
p(101) &= 102.
\end{align*}Find $p(102).$

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