You are given an integer n. In one move, you can either multiply n by two or divide n by 6 (if it is divisible by 6 without the remainder).

Your task is to find the minimum number of moves needed to obtain 1 from n or determine if it's impossible to do that.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow. 

The only line of the test case contains one integer n (1 ≤ n ≤ 10^9).

Output

For each test case, print the answer — the minimum number of moves needed to obtain 1 from n if it's possible to do that or -1 if it's impossible to obtain 1 from n.

Example

Input


7
1
2
3
12
12345
15116544
387420489


Output


0
-1
2
-1
-1
12
36

Note

Consider the sixth test case of the example. The answer can be obtained by the following sequence of moves from the given integer 15116544:

  1. Divide by 6 and get 2519424; 
  2. divide by 6 and get 419904; 
  3. divide by 6 and get 69984; 
  4. divide by 6 and get 11664; 
  5. multiply by 2 and get 23328; 
  6. divide by 6 and get 3888; 
  7. divide by 6 and get 648; 
  8. divide by 6 and get 108; 
  9. multiply by 2 and get 216; 
  10. divide by 6 and get 36; 
  11. divide by 6 and get 6; 
  12. divide by 6 and get 1. 
1
999838675

-1

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define ll long long
#define ld long double
#define fastio ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
#define pb push_back
#define pf push_front
#define mp make_pair
#define mod 1000000007
#define vll vector<ll>
#define inf 1e12
#define pll pair<ll,ll>
#define PI acos(-1)
#define pr 998244353
#define endl '\n'
using namespace std;
using namespace __gnu_pbds; 
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
/*
ll power(ll x,ll y)
{
	ll res=1;
	while(y>0)
	{
		if(y%2==1)res=(res*x)%mod;
		y=y/2;
		x=(x*x)%mod;
	}
	return res;
}
 
ll inv(ll a) 
{
return power(a,mod-2);
}
ll fact[2000007];
ll inr[2000007];
void pre()
{
	fact[0]=1;
	for(ll i=1;i<=2000005;i++)fact[i]=(fact[i-1]*i)%mod;
	inr[0]=1;
	for(ll i=1;i<=2000005;i++)inr[i]=inv(fact[i]);
}
ll ncr(ll n,ll r)
{
	ll a=(fact[n]*inr[r])%mod;
	ll b=(a*inr[n-r])%mod;
	return b;
}
 
ll spf[10000050];
void sieve()
{
	for(ll i=2;i*i<=10000030;i++)
	{
		if(spf[i]==0)
		{
			for(ll j=i+i;j<=10000030;j+=i)
			{
				if(spf[j]==0)
				{
					spf[j]=i;
				}
			}
		}
	}
}
*/
 
int main()
{
	//pre();
	//sieve();
	int t=1;
	cin>>t;	
	while(t--)
	{
		ll n;
		cin>>n;
		ll p2=0,p3=0;
		if(n==1)
		{
			cout<<"0"<<endl;
			continue;
		}
		if(n%6==0)
		{
			cout<<n/6<<endl;
			continue;
		}
		while(n%2==0)
		{
			p2++;
			n=n/2;
		}
		while(n%3==0)
		{
			p3++;
			n=n/3;
		}
		if(n!=1)
		{
			cout<<"-1";
		}
		else if(p2>=p3)cout<<"-1";
		else cout<<2*p3-p2;
		cout<<endl;
	}
}