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>
using namespace std;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
bool prime[10000001];
long long P = 1000000007 - 2;
long long factorial[10000001] = {0};
long long powerFunction(long long x, long long y) {
  long long res = 1;
  long long p = 1000000007;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void sieveFunction(long long maxLimit) {
  memset(prime, true, sizeof(prime));
  prime[0] = prime[1] = false;
  for (long long i = 2; maxLimit >= i; i++) {
    if (prime[i]) {
      for (long long j = 2 * i; maxLimit >= j; j += i) prime[j] = false;
    }
  }
}
void factorialFunction(long long maxLimit) {
  factorial[0] = 1;
  for (long long i = 1; i <= maxLimit; i++)
    factorial[i] = (factorial[i - 1] * i) % 1000000007;
  return;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long twos = 0;
    long long temp = n;
    while (temp % 2 == 0) {
      temp /= 2;
      twos++;
    }
    long long three = 0;
    temp = n;
    while (temp % 3 == 0) {
      temp /= 3;
      three++;
    }
    if (twos > three) {
      cout << -1 << "\n";
    } else {
      long long ops = three - twos;
      ops += three;
      cout << ops << "\n";
    }
  }
}
