super_log （广义欧拉降幂）（2019南京网络赛）

题目：

In Complexity theory, some functions are nearly O(1)O(1), but it is greater then O(1)O(1). For example, the complexity of a typical disjoint set is O(nα(n))O(nα(n)). Here α(n)α(n) is Inverse Ackermann Function, which growth speed is very slow. So in practical application, we often assume α(n) \le 4α(n)≤4.

However O(α(n))O(α(n)) is greater than O(1)O(1), that means if nn is large enough, α(n)α(n) can greater than any constant value.

Now your task is let another slowly function log*log∗ xx reach a constant value bb. Here log*log∗ is iterated logarithm function, it means “the number of times the logarithm function iteratively applied on xx before the result is less than logarithm base aa”.

Formally, consider a iterated logarithm function log_{a}^*loga∗​

Find the minimum positive integer argument xx, let log_{a}^* (x) \ge bloga∗​(x)≥b. The answer may be very large, so just print the result xx after mod mm.

Input

The first line of the input is a single integer T(T\le 300)T(T≤300) indicating the number of test cases.

Each of the following lines contains 33 integers aa, bb and mm.

1 \le a \le 10000001≤a≤1000000

0 \le b \le 10000000≤b≤1000000

1 \le m \le 10000001≤m≤1000000

Note that if a==1, we consider the minimum number x is 1.

Output

For each test case, output xx mod mm in a single line.

Hint

In the 4-th4−th query, a=3a=3 and b=2b=2. Then log_{3}^* (27) = 1+ log_{3}^* (3) = 2 + log_{3}^* (1)=3+(-1)=2 \ge blog3∗​(27)=1+log3∗​(3)=2+log3∗​(1)=3+(−1)=2≥b, so the output is 2727 mod 16 = 1116=11.

样例输入复制

5
2 0 3
3 1 2
3 1 100
3 2 16
5 3 233

样例输出复制

1
1
3
11
223

解题报告：严重怀疑这场比赛的出题方，读题实在是太困难了，看了半个小时才理解了要求解什么东西，一看就是一个广义欧拉降幂，然后就直接去写的之前的模板，求解无穷个2的幂次，
但是修修改改的过了样例，但是忘记了广义欧拉降幂的实现，疯狂wa，在比赛结束后找了好久才发现是没有去使用广义欧拉降幂的性质，忘记在该加欧拉（m）的地方进行加了，隔壁队的
数论手，给了一个能够实现这个功能的快速幂，稍微一修改就A了。这个需要掌握了，当时时间太紧，头脑已经混了。

ac代码：

 1 #include
 2 #include
 3 #include
 4 #include
 5 #include
 6 #include