[ICPC2021 Nanjing R] Xingqiu's Joke


## 题目描述 有 $T$ 个盒子,每盒子上有一个锁,锁上有两个整数 $a$ 和 $b$。你可以对这个锁做若干次以下 3 种操作: - $a$ 和 $b$ 分别减去 $1$ - $a$ 和 $b$ 分别增加 $1$ - $a$ 和 $b$ 分别除以它们共同的素数因子 如果 $a$ 或 $b$ 或两者都变为 $1$,盒子就会解锁。请你编写一个程序,计算每个盒子的锁打开的最少步骤数量。 ## 输入格式 第一行输入一个整数 $T(1≤T≤300)$。 接下来 $T$ 行,每行输入 $a$ 和 $b$,表示每个盒子的锁的信息。 ## 输出格式 共输出 $T$ 行,每行输出对应盒子解锁的最少步骤。


Once again, Xingqiu hides Chongyun's ice cream into a box with a strange lock. Liyue's summer has been always very hot and Chongyun suffers more because of his excessive yang (positive) energy, so he needs that ice cream desperately. ![](https://cdn.luogu.com.cn/upload/image_hosting/2dtcr426.png) There are two integers $a$ and $b$ on the lock. Chongyun can perform the following three types of operations any number of times: - Minus $1$ from both $a$ and $b$; - Plus $1$ to both $a$ and $b$; - Divide both $a$ and $b$ by one of their common $\textbf{prime}$ factor (that is to say, divide them by a $\textbf{prime}$ $g$ where $a$ and $b$ are both divisible by $g$). The box will be unlocked if either $a$ or $b$ or both become $1$. To help Chongyun gets the ice cream back as quickly as possible, please tell him the minimum number of operations needed to unlock the box.



There are multiple test cases. The first line of the input contains an integer $T$ ($1 \le T \le 300$) indicating the number of test cases. For each test case: The first and only line contains two integers $a$ and $b$ ($1 \le a, b \le 10^9$, $a \ne b$).


For each test case output one line containing one integer indicating the minimum number of operations to make $a$ or $b$ or both equal $1$.


输入样例 #1

4 7
9 8
32 84
11 35
2 1

输出样例 #1



For the first sample test case, the optimal way is $(4, 7) \rightarrow (3, 6) \rightarrow (1, 2)$. For the second sample test case, the optimal way is to apply the first type of operation $7$ times. For the third sample test case, the optimal way is $(32, 84) \rightarrow (16, 42) \rightarrow (15, 41) \rightarrow (14, 40) \rightarrow (13, 39) \rightarrow (1, 3)$. For the fourth sample test case, the optimal way is $(11, 35) \rightarrow (12, 36) \rightarrow (6, 18) \rightarrow (2, 6) \rightarrow (1, 3)$.