Buckets of Water (noch nicht übersetzt)

There are 3 buckets labelled $S$ (small) of 3 litres, $M$ (medium) of 5 litres and $L$ (large) of 8 litres.
Initially $S$ and $M$ are full of water and $L$ is empty. By pouring water between the buckets exactly one litre of water can be measured.
Since there is no other way to measure, once a pouring starts it cannot stop until either the source bucket is empty or the destination bucket is full.
At least four pourings are needed to get one litre:

After these operations, there is exactly one litre in bucket $S$.

In general the sizes of the buckets $S, M, L$ are $a$, $b$, $a + b$ litres, respectively. Initially $S$ and $M$ are full and $L$ is empty. If the above rule of pouring still applies and $a$ and $b$ are two coprime positive integers with $a\leq b$ then it is always possible to measure one litre in finitely many steps.

Let $P(a,b)$ be the minimal number of pourings needed to get one litre. Thus $P(3,5)=4$.
Also, $P(7, 31)=20$ and $P(1234, 4321)=2780$.

Find the sum of $P(2^{p^5}-1, 2^{q^5}-1)$ for all pairs of prime numbers $p,q$ such that $p < q < 1000$.
Give your answer modulo $1\,000\,000\,007$.