Codeforces Round #565 (Div. 3) A. Divide it!

Codeforces Round #565

 

A. Divide it!

You are given an integer n.

You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:

1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2*n/3 if n is divisible by 3;
3. Replace n with 4*n/5 if n is divisible by 5.

For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.

Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.

You have to answer q independent queries.

Input

The first line of the input contains one integer q (1≤q≤1000) — the number of queries.

The next q lines contain the queries. For each query you are given the integer number n (1≤n≤10^18).

Output

Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.

Example

 

input

7
1
10
25
30
14
27
1000000000000000000

output

0
4
6
6
-1
6
72



题意：这题意思是给你一个数，让你进行三种操作，1:能被2整除，变成n/2 。2：能被3整除，变成2*n/3 。3：能被5整除，变成4*n/5；
问你变到1的最少操作次数，如果不能变成1，则输出-1。

思路：直接暴力就可以了......

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
using namespace std;
#define ll long long
//const int maxn=;
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int T;
    cin>>T;
    ll int num;
    while(T--)
    {
        cin>>num;
        if(num==1)
        {
            cout<<0<<endl;
            continue;
        }
        ll int ans=0;
        while(num!=1)
        {
            if(num%2==0)
                num>>=1;
            else if(num%3==0)
            {
                num/=3;
                num<<=1;
            }
            else if(num%5==0)
            {
                num/=5;
                num<<=2;
            }
            else
            {
                ans=-1;
                break;
            }
            ans++;
        }
        cout<<ans<<endl;
    } 
    return 0;
}