https://codeforces.com/contest/1176/problem/A

题意：采取三种操作，使得n变成1的步骤数；否则-1

题解：模拟

1、n分解成质数乘积；

2、ans=2的个数+3的个数*2+5的个数*3；

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    scanf("%I64d",&t);
    while(t--){
        scanf("%I64d",&n);
        ans=0;
        while(n%2==0)n=n/2,ans++;
        while(n%3==0)n=n/3,ans+=2;
        while(n%5==0)n=n/5,ans+=3;
        cout<<(n==1?ans:-1)<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}