1933C · Turtle Fingers: Count the Values of k

Problem: You are given three positive integers $a$, $b$ and $l$ ($a,b,l>0$).

It can be shown that there always exists a way to choose non-negative (i.e. $\ge 0$) integers $k$, $x$, and $y$ such that $l = k \cdot a^x \cdot b^y$.

Your task is to find the number of distinct possible values of $k$ across all such ways.

Input Format: The first line contains the integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The following $t$ lines contain three integers, $a$, $b$ and $l$ ($2 \le a, b \le 100$, $1 \le l \le 10^6$) — description of a test case.

Output Format: Output $t$ lines, with the $i$-th ($1 \le i \le t$) line containing an integer, the answer to the $i$-th test case.

Note: In the first test case, $a=2, b=5, l=20$. The possible values of $k$ (and corresponding $x,y$) are as follows:

• Choose $k = 1, x = 2, y = 1$. Then $k \cdot a^x \cdot b^y = 1 \cdot 2^2 \cdot 5^1 = 20 = l$.
• Choose $k = 2, x = 1, y = 1$. Then $k \cdot a^x \cdot b^y = 2 \cdot 2^1 \cdot 5^1 = 20 = l$.
• Choose $k = 4, x = 0, y = 1$. Then $k \cdot a^x \cdot b^y = 4 \cdot 2^0 \cdot 5^1 = 20 = l$.
• Choose $k = 5, x = 2, y = 0$. Then $k \cdot a^x \cdot b^y = 5 \cdot 2^2 \cdot 5^0 = 20 = l$.
• Choose $k = 10, x = 1, y = 0$. Then $k \cdot a^x \cdot b^y = 10 \cdot 2^1 \cdot 5^0 = 20 = l$.
• Choose $k = 20, x = 0, y = 0$. Then $k \cdot a^x \cdot b^y = 20 \cdot 2^0 \cdot 5^0 = 20 = l$.

In the second test case, $a=2, b=5, l=21$. Note that $l = 21$ is not divisible by either $a = 2$ or $b = 5$. Therefore, we can only set $x = 0, y = 0$, which corresponds to $k = 21$.

In the third test case, $a=4, b=6, l=48$. The possible values of $k$ (and corresponding $x,y$) are as follows:

• Choose $k = 2, x = 1, y = 1$. Then $k \cdot a^x \cdot b^y = 2 \cdot 4^1 \cdot 6^1 = 48 = l$.
• Choose $k = 3, x = 2, y = 0$. Then $k \cdot a^x \cdot b^y = 3 \cdot 4^2 \cdot 6^0 = 48 = l$.
• Choose $k = 8, x = 0, y = 1$. Then $k \cdot a^x \cdot b^y = 8 \cdot 4^0 \cdot 6^1 = 48 = l$.
• Choose $k = 12, x = 1, y = 0$. Then $k \cdot a^x \cdot b^y = 12 \cdot 4^1 \cdot 6^0 = 48 = l$.
• Choose $k = 48, x = 0, y = 0$. Then $k \cdot a^x \cdot b^y = 48 \cdot 4^0 \cdot 6^0 = 48 = l$.