---

Problem 758

---

Buckets of Water

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:

$$(3,5,0)\stackrel{M\to L}{\to }(3,0,5)\stackrel{S\to M}{\to }(0,3,5)\stackrel{L\to S}{\to }(3,3,2)\stackrel{S\to M}{\to }(1,5,2)$$

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 $a\le b$ 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 $1000000007$.

---

水桶倒水

现有三个水桶，标记$S$的小水桶可以装$3$升水，标记$M$的中水桶可以装$5$升水，标记$L$的大水桶可以装$8$升水。
一开始，小水桶和中水桶均装满了水，而大水桶则是空的。通过在水桶间互相倒水，我们希望量出恰好一升水。由于我们缺乏其他的测量手段，在倒水时只能要么将倒出的水桶倒空要么将倒入的水桶装满。
为了量出恰好一升水，至少需要倒四次：

$$(3,5,0)\stackrel{M\to L}{\to }(3,0,5)\stackrel{S\to M}{\to }(0,3,5)\stackrel{L\to S}{\to }(3,3,2)\stackrel{S\to M}{\to }(1,5,2)$$

经过这些操作后，在小水桶中恰好有一升水。

考虑一般情况，即小、中、大水桶的容积分别为$a$、$b$和$a+b$升。一开始小水桶和中水桶均装满了水而大水桶是空的。采用相同的倒水规则，若$a$和$b$是两个互质的正整数且满足$a\le b$，那么在有限步内总是能够量出恰好一升水。

记$P(a,b)$为量出一升水所需的最少倒水次数，因此$P(3,5)=4$。
此外，$P(7,31)=20$，$P(1234,4321)=2780$。

对于所有满足$p<q<1000$的素数对$p,q$，求$P(2^{p^5}-1,2^{q^5}-1)$之和，并将你的答案对$1000000007$取余。

---