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
题解：求a^a^...(b次)%n的结果。因为n与a不一定互质，所以要利用广义欧拉定理进行降幂。
AC代码：
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