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>
  #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
  using namespace std;
  template <class T>
  using rque = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  bool chmin(T &a, const T &b) {
    if (b < a) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  bool chmax(T &a, const T &b) {
    if (b > a) {
      a = b;
      return 1;
    }
    return 0;
  }
  long long gcd(long long a, long long b) {
    if (a == 0) return b;
    if (b == 0) return a;
    long long cnt = a % b;
    while (cnt != 0) {
      a = b;
      b = cnt;
      cnt = a % b;
    }
    return b;
  }
  long long extGCD(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    long long d = extGCD(b, a % b, y, x);
    y -= a / b * x;
    return d;
  }
  struct UnionFind {
    vector<long long> data;
    int num;
    UnionFind(int sz) {
      data.assign(sz, -1);
      num = sz;
    }
    bool unite(int x, int y) {
      x = find(x), y = find(y);
      if (x == y) return (false);
      if (data[x] > data[y]) swap(x, y);
      data[x] += data[y];
      data[y] = x;
      num--;
      return (true);
    }
    int find(int k) {
      if (data[k] < 0) return (k);
      return (data[k] = find(data[k]));
    }
    long long size(int k) { return (-data[find(k)]); }
    bool same(int x, int y) { return find(x) == find(y); }
  };
  template <int mod>
  struct ModInt {
    int x;
    ModInt() : x(0) {}
    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
    ModInt &operator+=(const ModInt &p) {
      if ((x += p.x) >= mod) x -= mod;
      return *this;
    }
    ModInt &operator-=(const ModInt &p) {
      if ((x += mod - p.x) >= mod) x -= mod;
      return *this;
    }
    ModInt &operator*=(const ModInt &p) {
      x = (int)(1LL * x * p.x % mod);
      return *this;
    }
    ModInt &operator/=(const ModInt &p) {
      *this *= p.inverse();
      return *this;
    }
    ModInt operator-() const { return ModInt(-x); }
    ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
    ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
    ModInt &operator++() { return *this += ModInt(1); }
    ModInt operator++(int) {
      ModInt tmp = *this;
      ++*this;
      return tmp;
    }
    ModInt &operator--() { return *this -= ModInt(1); }
    ModInt operator--(int) {
      ModInt tmp = *this;
      --*this;
      return tmp;
    }
    ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
    ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
    bool operator==(const ModInt &p) const { return x == p.x; }
    bool operator!=(const ModInt &p) const { return x != p.x; }
    ModInt inverse() const {
      int a = x, b = mod, u = 1, v = 0, t;
      while (b > 0) {
        t = a / b;
        swap(a -= t * b, b);
        swap(u -= t * v, v);
      }
      return ModInt(u);
    }
    ModInt pow(int64_t n) const {
      ModInt ret(1), mul(x);
      while (n > 0) {
        if (n & 1) ret *= mul;
        mul *= mul;
        n >>= 1;
      }
      return ret;
    }
    friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x; }
    friend istream &operator>>(istream &is, ModInt &a) {
      int64_t t;
      is >> t;
      a = ModInt<mod>(t);
      return (is);
    }
    static int get_mod() { return mod; }
  };
  long long mpow2(long long x, long long n, long long mod) {
    long long ans = 1;
    while (n != 0) {
      if (n & 1) ans = ans * x % mod;
      x = x * x % mod;
      n = n >> 1;
    }
    return ans;
  }
  long long modinv2(long long a, long long mod) {
    long long b = mod, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    u %= mod;
    if (u < 0) u += mod;
    return u;
  }
  constexpr int mod = 998244353;
  using mint = ModInt<mod>;
  mint mpow(mint x, long long n) {
    mint ans = 1;
    while (n != 0) {
      if (n & 1) ans *= x;
      x *= x;
      n = n >> 1;
    }
    return ans;
  }
  int main() {
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    cout << fixed << setprecision(15);
    int T;
    cin >> T;
    while (T--) {
      int n;
      cin >> n;
      vector<int> a(n);
      for (long long i = 0; i < (n); i++) cin >> a[i];
      int g = 0;
      for (long long t = 0; t < (30); t++) {
        int cnt = 0;
        for (long long i = 0; i < (n); i++) cnt += (a[i] >> t & 1);
        g = gcd(g, cnt);
      }
      vector<int> ans;
      for (long long i = 1; i < (n + 1); i++)
        if (gcd(g, i) == i) ans.emplace_back(i);
      for (long long i = 0; i < (ans.size()); i++)
        cout << ans[i] << (i + 1 == ans.size() ? '\n' : ' ');
    }
  }
  