A positive integer $n$ is considered $B$-trivisible if the sum of all different prime factors of $n$ which are not larger than $B$ is divisible by $3$.

For example, $175=5^2\cdot 7$ is $10$-trivisible because $5+7=12$ which is divisible by $3$. Similarly, $175$ is $4$-trivisible because all primes dividing $175$ are larger than $4$, and the empty summation $0$ is divisible by $3$.
On the other hand, $175$ is not $6$-trivisible because the sum of relevant primes is $5$ which is not divisible by $3$.

Let $F(N,B)$ be the number of $B$-trivisible integers not larger than $N$.

For example, $F(10,4)=5$, the $4$-trivisible numbers being $1,3,5,7,9$.
You are also given $F(10,10)=3$ and $F(100,10)=41$.

Find $F({10}^{18},120)$.