Eccoci.

Anton and Bertrand love to play three pile Nim.
However, after a lot of games of Nim they got bored and changed the rules somewhat.
They may only take a number of stones from a pile that is a proper divisora proper divisor of $n$ is a divisor of $n$ smaller than $n$ of the number of stones present in the pile.
E.g. if a pile at a certain moment contains $24$ stones they may take only $1,2,3,4,6,8$ or $12$ stones from that pile.
So if a pile contains one stone they can't take the last stone from it as $1$ isn't a proper divisor of $1$.
The first player that can't make a valid move loses the game.
Of course both Anton and Bertrand play optimally.

The triple $(a, b, c)$ indicates the number of stones in the three piles.
Let $S(n)$ be the number of winning positions for the next player for $1 \le a, b, c \le n$.
$S(10) = 692$ and $S(100) = 735494$.

Find $S(123456787654321)$ modulo $1234567890$.