description: You are given array a_1, a_2, …, a_n, consisting of non-negative integers.
  
  Let's define operation of "elimination" with integer parameter k (1 ≤ k ≤ n) as follows:
  
    * Choose k distinct array indices 1 ≤ i_1 < i_2 < … < i_k ≤ n. 
    * Calculate x = a_{i_1} ~ \& ~ a_{i_2} ~ \& ~ … ~ \& ~ a_{i_k}, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND) (notes section contains formal definition). 
    * Subtract x from each of a_{i_1}, a_{i_2}, …, a_{i_k}; all other elements remain untouched. 
  
  
  
  Find all possible values of k, such that it's possible to make all elements of array a equal to 0 using a finite number of elimination operations with parameter k. It can be proven that exists at least one possible k for any array a.
  
  Note that you firstly choose k and only after that perform elimination operations with value k you've chosen initially.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
  
  The first line of each test case contains one integer n (1 ≤ n ≤ 200 000) — the length of array a.
  
  The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{30}) — array a itself.
  
  It's guaranteed that the sum of n over all test cases doesn't exceed 200 000.
  
  Output
  
  For each test case, print all values k, such that it's possible to make all elements of a equal to 0 in a finite number of elimination operations with the given parameter k.
  
  Print them in increasing order.
  
  Example
  
  Input
  
  
  5
  4
  4 4 4 4
  4
  13 7 25 19
  6
  3 5 3 1 7 1
  1
  1
  5
  0 0 0 0 0
  
  
  Output
  
  
  1 2 4
  1 2
  1
  1
  1 2 3 4 5
  
  Note
  
  In the first test case:
  
    * If k = 1, we can make four elimination operations with sets of indices \{1\}, \{2\}, \{3\}, \{4\}. Since \& of one element is equal to the element itself, then for each operation x = a_i, so a_i - x = a_i - a_i = 0. 
    * If k = 2, we can make two elimination operations with, for example, sets of indices \{1, 3\} and \{2, 4\}: x = a_1 ~ \& ~ a_3 = a_2 ~ \& ~ a_4 = 4 ~ \& ~ 4 = 4. For both operations x = 4, so after the first operation a_1 - x = 0 and a_3 - x = 0, and after the second operation — a_2 - x = 0 and a_4 - x = 0. 
    * If k = 3, it's impossible to make all a_i equal to 0. After performing the first operation, we'll get three elements equal to 0 and one equal to 4. After that, all elimination operations won't change anything, since at least one chosen element will always be equal to 0. 
    * If k = 4, we can make one operation with set \{1, 2, 3, 4\}, because x = a_1 ~ \& ~ a_2 ~ \& ~ a_3 ~ \& ~ a_4 = 4. 
  
  
  
  In the second test case, if k = 2 then we can make the following elimination operations:
  
    * Operation with indices \{1, 3\}: x = a_1 ~ \& ~ a_3 = 13 ~ \& ~ 25 = 9. a_1 - x = 13 - 9 = 4 and a_3 - x = 25 - 9 = 16. Array a will become equal to [4, 7, 16, 19]. 
    * Operation with indices \{3, 4\}: x = a_3 ~ \& ~ a_4 = 16 ~ \& ~ 19 = 16. a_3 - x = 16 - 16 = 0 and a_4 - x = 19 - 16 = 3. Array a will become equal to [4, 7, 0, 3]. 
    * Operation with indices \{2, 4\}: x = a_2 ~ \& ~ a_4 = 7 ~ \& ~ 3 = 3. a_2 - x = 7 - 3 = 4 and a_4 - x = 3 - 3 = 0. Array a will become equal to [4, 4, 0, 0]. 
    * Operation with indices \{1, 2\}: x = a_1 ~ \& ~ a_2 = 4 ~ \& ~ 4 = 4. a_1 - x = 4 - 4 = 0 and a_2 - x = 4 - 4 = 0. Array a will become equal to [0, 0, 0, 0]. 
  
  
  
  Formal definition of bitwise AND:
  
  Let's define bitwise AND (\&) as follows. Suppose we have two non-negative integers x and y, let's look at their binary representations (possibly, with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of number x, and y_i is the i-th bit of number y. Let r = x ~ \& ~ y is a result of operation \& on number x and y. Then binary representation of r will be r_k ... r_2 r_1 r_0, where:
  
  $$$ r_i = \begin{cases} 1, ~ if ~ x_i = 1 ~ and ~ y_i = 1 \\\ 0, ~ if ~ x_i = 0 ~ or ~ y_i = 0 \end{cases} $$$
solution: #include <bits/stdc++.h>
  using namespace std;
  using ll = long long;
  using ld = double;
  using pi = pair<int, int>;
  using pl = pair<ll, ll>;
  using pd = pair<ld, ld>;
  using cd = complex<ld>;
  using vcd = vector<cd>;
  using vi = vector<int>;
  using vl = vector<ll>;
  using vd = vector<ld>;
  using vs = vector<string>;
  using vpi = vector<pi>;
  using vpl = vector<pl>;
  using vpd = vector<pd>;
  template <class T>
  using min_pq = priority_queue<T, vector<T>, greater<T> >;
  template <class T>
  inline int ckmin(T& a, const T& val) {
    return val < a ? a = val, 1 : 0;
  }
  template <class T>
  inline int ckmax(T& a, const T& val) {
    return a < val ? a = val, 1 : 0;
  }
  template <class T>
  void remDup(vector<T>& v) {
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), end(v));
  }
  constexpr int pct(int x) { return __builtin_popcount(x); }
  constexpr int bits(int x) { return x == 0 ? 0 : 31 - __builtin_clz(x); }
  constexpr int p2(int x) { return 1 << x; }
  constexpr int msk2(int x) { return p2(x) - 1; }
  ll ceilDiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
  ll floorDiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
  void setPrec(int x) { cout << fixed << setprecision(x); }
  string to_string(char c) { return string(1, c); }
  string to_string(const char* s) { return (string)s; }
  string to_string(string s) { return s; }
  string to_string(bool b) { return to_string((int)b); }
  template <class T>
  string to_string(complex<T> c) {
    stringstream ss;
    ss << c;
    return ss.str();
  }
  template <class T>
  using V = vector<T>;
  string to_string(V<bool> v) {
    string res = "{";
    for (int i = (0); i <= (int((v).size()) - 1); i++) res += char('0' + v[i]);
    res += "}";
    return res;
  }
  template <size_t sz>
  string to_string(bitset<sz> b) {
    string res = "";
    for (int i = (0); i <= (int((b).size()) - 1); i++) res += char('0' + b[i]);
    return res;
  }
  template <class T, class U>
  string to_string(pair<T, U> p);
  template <class T>
  string to_string(T v) {
    bool fst = 1;
    string res = "";
    for (const auto& x : v) {
      if (!fst) res += " ";
      fst = 0;
      res += to_string(x);
    }
    return res;
  }
  template <class T, class U>
  string to_string(pair<T, U> p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  }
  template <class T>
  void pr(T x) {
    cout << to_string(x);
  }
  template <class T, class... U>
  void pr(const T& t, const U&... u) {
    pr(t);
    pr(u...);
  }
  void ps() { pr("\n"); }
  template <class T, class... U>
  void ps(const T& t, const U&... u) {
    pr(t);
    if (sizeof...(u)) pr(" ");
    ps(u...);
  }
  void DBG() { cerr << "]" << endl; }
  template <class T, class... U>
  void DBG(const T& t, const U&... u) {
    cerr << to_string(t);
    if (sizeof...(u)) cerr << ", ";
    DBG(u...);
  }
  const ld PI = acos(-1.0);
  const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
  const ld EPS = 1e-9;
  const ll MODBASE = 1000000007LL;
  const int INF = 0x3f3f3f3f;
  const int MAXN = 110;
  const int MAXM = 1000;
  const int MAXK = 16;
  const int MAXQ = 200010;
  int f[30], n;
  int gcd(int a, int b) {
    while (b) {
      a %= b;
      swap(a, b);
    }
    return a;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    int te;
    cin >> te;
    while (te--) {
      cin >> n;
      memset(f, 0, sizeof(f));
      for (int i = (1); i <= (n); i++) {
        int x;
        cin >> x;
        for (int j = (29); j >= (0); j--)
          if (x & (1 << j)) f[j]++;
      }
      int g = 0;
      for (int i = (0); i <= (29); i++) g = gcd(g, f[i]);
      if (g == 0) {
        for (int i = (1); i <= (n); i++) cout << i << " ";
        cout << "\n";
      } else {
        for (int i = (1); i <= (g); i++)
          if (g % i == 0) cout << i << ' ';
        cout << "\n";
      }
    }
    return 0;
  }
  