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;
long long int min(long long int a, long long int b) { return a < b ? a : b; }
long long int min(long long int a, long long int b, long long int c) {
  a = min(a, b);
  a = min(a, c);
  return a;
}
long long int max(long long int a, long long int b) { return a > b ? a : b; }
long long int max(long long int a, long long int b, long long int c) {
  a = max(a, b);
  a = max(a, c);
  return a;
}
long long int dsum(long long int n) {
  long long int ans = 0;
  while (n) {
    ans += (n % 10);
    n /= 10;
  }
  return ans;
}
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int power(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  if (b == 1) return a % m;
  long long int res = 1;
  a = a % m;
  while (b) {
    if (b & 1) {
      res = (res * a) % m;
    }
    a = ((a % m) * (a % m)) % m;
    b >>= 1;
  }
  return res;
}
long long int Ceil(long long int a, long long int b) {
  long long int res = a / b;
  if (a % b != 0) res++;
  return res;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &t : v) is >> t;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (const T &t : v) {
    os << t << " ";
  }
  os << '\n';
  return os;
}
long long int dxx[] = {1, 1, 0, -1, -1, -1, 0, 1};
long long int dyy[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long int dx[] = {1, 0, -1, 0};
long long int dy[] = {0, 1, 0, -1};
struct pll {
  long long int x, y;
};
bool comp(pll a, pll b) {
  if (a.x == b.x)
    return a.y < b.y;
  else
    return a.x < b.x;
}
void solve(long long int the) {
  char ch;
  double aa, bb, cc, dd, xx;
  string s1, s2, s3, str;
  long long int i, j, k, a, b, c, d, n, m, l, r, x, y, z, low, high, mid,
      sum = 0, ans = 0, temp, t;
  cin >> n;
  if (n == 0) {
    cout << -1 << endl;
    return;
  }
  if (n == 1) {
    cout << 0 << endl;
    return;
  }
  if (n == 2) {
    cout << -1 << endl;
    return;
  }
  if (n == 3) {
    cout << 2 << endl;
    return;
  }
  if (n == 4 || n == 5) {
    cout << -1 << endl;
    return;
  }
  ans = 0;
  if (n & 1) {
    n = 2 * n;
    ans = 1;
  }
  while (n % 6 == 0) {
    n /= 6;
    ans++;
    if (n == 1) break;
    if (n & 1) {
      n = 2 * n;
      ans++;
    }
    if (n % 6 != 0) break;
  }
  if (n != 1)
    cout << -1 << endl;
  else
    cout << ans << endl;
}
int main() {
  long long int t = 1, c = 1;
  scanf("%lld", &t);
  while (t--) {
    solve(c);
    c++;
  }
  return 0;
}
