两数相除，判断小数位是否有限位

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≤n≤105

) — the number of queries.

Next n

lines contain queries, one per line. Each line contains three integers p, q, and b (0≤p≤1018, 1≤q≤1018, 2≤b≤1018). 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

26 12 104 3 10

Output

Copy

FiniteInfinite

Input

Copy

41 1 29 36 24 12 33 5 4

Output

Copy

FiniteFiniteFiniteInfinite

Note

612=12=0,510

43=1,(3)10

936=14=0,012

412=13=0,13

题意 ： 给两个数，以及一个所要求的进制，询问是否在此进制下，相除后是否为有限位小数

思路分析 ： 比赛的时候想了一个模拟，不过没有去写，旁边的大佬一直再和说，会超时，会超时，然后赛后才知道是用这样的，比如 10进制下，判断两数是否整除，只需要让除数已知除以 2 或者 5，当可以除到 1 的时候，代表是可以整除的，如果是换成任意进制下的一个数，则需要去除当前的数和 b进制的约数。

代码示例 ：

#define ll long long

ll gcd(ll a, ll b){
    return b == 0?a:gcd(b, a%b);
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ll t;
    ll p, q, b;

    cin >> t;
    while(t--){
        scanf("%lld%lld%lld", &p, &q, &b);
        ll g = gcd(p, q);
        ll x = q / g;
        ll f = gcd(b, x);

        while(1){
            if (f == 1) break;
            while(x%f == 0) x /= f;
            f = gcd(b, x);
        }
        if (x == 1) printf("Finite\n");
        else printf("Infinite\n");
    }
    return 0;
}

原文地址：https://www.cnblogs.com/ccut-ry/p/9047523.html