Codeforces Round #483 (Div. 2) C. Finite or not?

C. Finite or not?

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given several queries. Each query consists of three integers pp, qq and bb. You need to answer whether the result of p/qp/q in notation with base bb is a finite fraction.

A fraction in notation with base bb is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.

Input

The first line contains a single integer nn (1≤n≤1051≤n≤105) — the number of queries.

Next nn lines contain queries, one per line. Each line contains three integers pp, qq, and bb (0≤p≤10180≤p≤1018, 1≤q≤10181≤q≤1018, 2≤b≤10182≤b≤1018). All numbers are given in notation with base 1010.

Output

For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.

Examples

input

Copy

2
6 12 10
4 3 10

output

Copy

Finite
Infinite

input

Copy

4
1 1 2
9 36 2
4 12 3
3 5 4

output

Copy

Finite
Finite
Finite
Infinite

Note

612=12=0,510612=12=0,510

43=1,(3)1043=1,(3)10

936=14=0,012936=14=0,012

412=13=0,13412=13=0,13

这题，看这个分数在b进制下是否为无限循环小数。

一个分数是否为无限循环小数取决于分母。

于是想到这题一定是考虑 分母q 和 进制b 的关系，

于是就是想办法  把分母化为1就不是无限循环小数

temp1 = gcd(q, b);

if (temp1 == 1) break;
while(q % temp1== 0) {
      q /= temp1;
}
while一定要加  不然会超时。

 #include<bits/stdc++.h>

 using namespace std;
 const int maxn = 3e5 + ;
 typedef long long LL;
 LL gcd(LL a, LL b ) {
     if (b == ) return a;
     return gcd(b, a % b);
 }
 int main() {
     int n;
     LL p, q, b;
     scanf("%d", &n);
     while(n--) {
         scanf("%lld%lld%lld", &p, &q, &b);
         LL g = gcd(p, q);
         p = p / g, q = q / g;
         LL temp = p / q;
         p = p - temp * q;
         if (!p) {
             printf("Finite\n");
             continue;
         }
         LL temp1;
         while() {
             temp1 = gcd(q, b);
             if (temp1 == ) break;
             while(q % temp1== ) {
                 q /= temp1;
             }
         }
         if (q ==  ) printf("Finite\n");
         else printf("Infinite\n");
     }
     return ;
 }