Ordinary Ordinals

Sets, sets, sets. Everything in math is just a set. Even the natural numbers can be represented as sets. For example, we can represent the the number $0$ as the empty set $\\{\\}$. The number $1$ can be represented as $\\{\\{\\}\\}$.

But what about $2$? Consider $\\{\\{\\},\\{\\{\\}\\}\\}$, the set containing both the empty set and $\\{\\{\\}\\}$. This is a nice choice for $2$ for two reasons: we have that $0$ and $1$ are elements of $2$ and we also have that $0$ and $1$ are subsets of $2$.

In general, for $N>0$ we can represent $N$ as the set $\\{0,1, \dots ,N-1\\}$ where we recursively apply the representations for $0,1, \dots ,N-1$. For example: \begin{eqnarray\*} 3 & = & \\{ 0, 1, 2\\} \\\ & = & \\{ \\{ \\} , \\{ 0\\} , \\{ 0, 1\\} \\} \\\ & = & \\{ \\{ \\} , \\{ \\{ \\} \\} , \\{ \\{ \\}, \\{ 0\\} \\} \\} \\\ & = & \\{ \\{ \\} , \\{ \\{ \\} \\} , \\{ \\{ \\} , \\{ \\{ \\} \\} \\} \\} \end{eqnarray\*}

Thus, for each $0≤i\<N$, $i$ is both a member of $N$ and a subset of $N$. Another nice feature is the size of the set representing $N$ is also $N$. But what is not so nice is the number of characters it takes to write such a set. Given a natural number $N≥0$, how many braces and commas are required to write out the set representing $N$ in the above manner?

More specifically, let $f(N)$ be the number of bracket and comma characters required to write out the set representing $N$. Since $f(N)$ can be quite large, your job is to determine $f(N)$ modulo some positive integer $M$.

Input consists of two integers $N$ ($0≤N<2^{63}$) and $M$ ($1≤M<2^{31}$) as described above.

Display the value of $f(N)$ reduced modulo $M$. That is, the remainder that would be left if you divided $f(N)$ by $M$.