OJ 1451 digital games

describe

Given an integer n.

You can use this number to do any of the following ( It could be zero ) frequency :

If n Can be 2 to be divisible by , Then use n/2 Instead of n;

If n Can be 3 to be divisible by , Then use 2n/3 Instead of n;

If n Can be 5 to be divisible by , Just use 4n/5 Instead of n.

for example , You can use the first operation to 30 Replace with 15, Using the second operation will 30 Replace with 20, Or use the third operation to 30 Replace with 24.

Your task is to find out from n Get in 1 Minimum number of steps required , Or this is impossible .

You must answer independently of q Query for .

Print the answer to each query in a new row . If you can't get from n Get in 1, Then print -1. otherwise , The minimum number of steps required for printing .

Input
The first line of input contains an integer q(1≤q≤1000)—— Number of queries .

Next q The row contains the query . For each query , There is an integer n(1≤n≤1018).

Output
Print the answer of each query in each row . If you can't get from n Get in 1, Then print -1. otherwise , The minimum number of steps required for printing .
Digital games
describe

Given an integer n.

You can use this number to do any of the following ( It could be zero ) frequency :

If n Can be 2 to be divisible by , Then use n/2 Instead of n;

If n Can be 3 to be divisible by , Then use 2n/3 Instead of n;

If n Can be 5 to be divisible by , Just use 4n/5 Instead of n.

for example , You can use the first operation to 30 Replace with 15, Using the second operation will 30 Replace with 20, Or use the third operation to 30 Replace with 24.

Your task is to find out from n Get in 1 Minimum number of steps required , Or this is impossible .

You must answer independently of q Query for .

Print the answer to each query in a new row . If you can't get from n Get in 1, Then print -1. otherwise , The minimum number of steps required for printing .

Input
The first line of input contains an integer q(1≤q≤1000)—— Number of queries .

Next q The row contains the query . For each query , There is an integer n(1≤n≤1018).

Output
Print the answer of each query in each row . If you can't get from n Get in 1, Then print -1. otherwise , The minimum number of steps required for printing .

sample input 1 

7
1
10
25
30
14
27
1000000000000000000
sample output 1

0
4
6
6
-1
6
72

The problem requires that the data be changed into 1, If a number changes many times and becomes a non Can be 2,3,5 The number divided by an integer, then this data cannot become 1, Otherwise, the number of steps may be less , Then the reduction should also be as large as possible , Cut from small to large 1/2,2/3,4/5, Then use this rule to analogy .

#include <iostream>

using namespace std;
long long MIN;
void doit (long long a,int cont)
{
    if(a%2==0)
    {
        if(a/2==1)
        {
            if(MIN>cont)
                MIN=cont;
        }
        else
            doit(a/2,cont+1);
    }
    else if(a%3==0)
    {
        if(2*a/3==1)
        {
            if(MIN>cont)
                MIN=cont;
        }
        else
            doit(2*a/3,cont+1);
    }
   else if(a%5==0)
    {
        if(4*a/5==1)
        {
            if(MIN>cont)
                MIN=cont;
        }
        else
            doit(4*a/5,cont+1);
    }
    else
        MIN=-1;

}
int main()
{
    long long n,a;
    while(cin>>n)
    {
        while(n--)
        {
            MIN=0x3f3f3f;
            cin>>a;
            if(a==1)
                cout<<"0"<<endl;
            else
            {
                doit(a,1);
                cout<<MIN<<endl;
            }

        }
    }
    return 0;
}