Description: 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 $$$k \cdot a^x \cdot b^y = 48 \cdot 4^0 \cdot 6^0 = 48 = l$$$.