69 人解决，92 人已尝试。
114 份提交通过，共有 403 份提交。
3.9 EMB 奖励。
单点时限: 2.0 sec
内存限制: 256 MB
Professor Shekhu has another problem for Akki today. He has given him three positive integers $A$, $N$ and $P$ and wants him to calculate the remainder when $A^{N!}$ is divided by $P$. As usual, $N!$ denotes the product of the first $N$ positive integers.
The first line of the input gives the number of test cases, $T$. $T$ lines follow. Each line contains three integers $A$, $N$ and $P$, as described above.
For each test case, output one line containing Case #x: y, where $x$ is the test case number (starting from 1) and $y$ is the answer.
2 2 1 2 3 3 2
Case #1: 0 Case #2: 1
In Sample Case #1, the answer is the remainder when $2^{1!} = 2$ is divided by 2, which is 0.
In Sample Case #2, the answer is the remainder when $3^{3!} = 36 = 729$ is divided by 2, which is 1.
69 人解决，92 人已尝试。
114 份提交通过，共有 403 份提交。
3.9 EMB 奖励。