The look and say sequence goes 1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ...
The sequence starts with 1 and all other members are obtained by describing the previous member in terms of consecutive digits.
It helps to do this out loud:
1 is 'one one' → 11
11 is 'two ones' → 21
21 is 'one two and one one' → 1211
1211 is 'one one, one two and two ones' → 111221
111221 is 'three ones, two twos and one one' → 312211
...

Define $A(n)$, $B(n)$ and $C(n)$ as the number of ones, twos and threes in the $n$'th element of the sequence respectively.
One can verify that $A(40)=31254$, $B(40)=20259$ and $C(40)=11625$.

Find $A(n)$, $B(n)$ and $C(n)$ for $n={10}^{12}$.
Give your answer modulo $2^{30}$ and separate your values for $A$, $B$ and $C$ by a comma.
E.g. for $n=40$ the answer would be 31254,20259,11625