Finite or not?(CF-#483 DIV2-C)(数论——GCD)

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 $p$, $q$ and $b$. You need to answer whether the result of $p/q$ in notation with base $b$ is a finite fraction.

A fraction in notation with base $b$ 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 $n$ ($1\le n\le {10}^5$) — the number of queries.

Next $n$ lines contain queries, one per line. Each line contains three integers $p$, $q$, and $b$ ($0\le p\le {10}^{18}$, $1\le q\le {10}^{18}$, $2\le b\le {10}^{18}$). All numbers are given in notation with base $10$.

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

$\frac{6}{12}=\frac{1}{2}=0,5_{10}$

$\frac{4}{3}=1,(3{)}_{10}$

$\frac{9}{36}=\frac{1}{4}=0,{01}_2$

$\frac{4}{12}=\frac{1}{3}=0,1_3$

首先恭喜杨哥哥在这次比赛中蓝名！！！，虽说有点（非常）妒忌，但是我也很开心，有人陪我打CF，还有人教我，

扫描二维码关注公众号，回复： 1014504 查看本文章

让我从黑名一步一步走出来，希望下一次就绿名。。。

题意：p/q   在b进制是否是一个  有限小数。

题解：这个题目有意思，

在杨哥哥的教导下，他说，首先对  分子分母约分，然后  再不断地对  b和q  /gcd（b,q）;

主要看   q==1 证明有限，反之就说明  无限。

但是这样很很容易超时，  所以   b不断减少，减少它们间的差距来达到  优化。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define gcd __gcd               //350(杨哥哥过的时间)
/*inline ll gcd(ll a,ll b){     //390
  while(b^=a^=b^=a%=b);
        return a;
}*/
/*inline ll gcd(ll a,ll b){     //420
    return a%b==0?b:gcd(b,a%b);
}*/
/*inline ll gcd(ll x,ll y){     //620
    ll i,j;
    if(x==0)return y;
    if(y==0)return x;
    for(i=0;0==(x&1);i++)x>>=1;
    for(j=0;0==(y&1);j++)y>>=1;
    if(j<i)i=j;
    while(1){
        if(x<y)x^=y,y^=x,x^=y;
        if(0==(x-=y))return y<<i;
        while(0==(x&1))x>>=1;
    }
}*/
int main()
{
    ll T,p,q,b;
    scanf("%lld",&T);
    while(T--){
        scanf("%lld%lld%lld",&p,&q,&b);
        ll t=gcd(p,q);
        q/=t;
        t=b;
        while(q!=1){
            t=gcd(q,t);
            q/=t;
            if(t==1)break;
        }
        if(q==1){
            printf("Finite\n");
        }else{
            printf("Infinite\n");
        }
    }
    return 0;
}