由于 $32768 = 2^{15}$，所以答案的上界是 15，猜想先对数字进行加，然后再乘，这样是最优的。假设某次乘操作之后，进行了两次加：$v \to 2v \to 2v + 1 \to 2v + 2$，此时可以替换成 $v \to v + 1 \to 2v + 2$，也就是乘操作之后，不能进行连续的两次加。乘操作之后再紧接着进行一次加操作不是最优的，因为我们得到的数字需要被 $2^{15}$ 整除，加操作破坏了整除性。因此只能先进行连续的加操作，然后是连续的乘操作。设加操作次数是 $i$，乘操作次数是 $j$，那么有 $(x + i) \times 2^{j} \bmod 2^{15} = 0$，$i, j \le 15$。

// Date: Sat Dec 16 20:56:23 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 400010, M = 32768;

int bfs(int x) {
  int res = 15;
  For1(i, 0, 15) {
    For1(j, 0, 15) {
      ll tmp = (x + i) * (1 << j) % M;
      if (tmp == 0) {
        res = min(res, i + j);
      }
    }
  }
  return res;
}

int a[N], b[N];

int main(void) {
#ifdef _DEBUG
  freopen("1661b.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  int n;
  while (cin >> n) {
    For1(i, 1, n) { cin >> a[i]; }

    For1(i, 1, n) { b[i] = bfs(a[i]); }

    For1(i, 1, n) { cout << b[i] << ' '; }
    cout << '\n';
  }

  return 0;
}