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;
  void speed() {
    cin.tie(0);
    cin.sync_with_stdio(0);
  }
  long long factorial(long long x) {
    long long res = 1;
    for (int i = 2; i <= x; i++) {
      res *= i;
    }
    return res;
  }
  long long remove_0(long long x) {
    long long res = 0, cnt = 1;
    while (x > 0) {
      int digit = x % 10;
      if (digit != 0) {
        res += (digit * cnt);
        cnt *= 10;
      }
      x /= 10;
    }
    return res;
  }
  bool prime[100005];
  void prime_numbers(int x) {
    memset(prime, true, sizeof(prime));
    int c = 0;
    prime[0] = prime[1] = false;
    for (int i = 2; i * i <= 100001; i++) {
      if (prime[i] == true) {
        for (int j = i * i; j <= 100001; j += i) {
          prime[j] = false;
        }
      }
    }
    return;
  }
  long long sum_binary(long long x) {
    long long s = 0;
    while (x > 0) {
      s += x % 2;
      x /= 2;
    }
    return s;
  }
  int binary_s(int a[], int l, int h, int k) {
    if (l <= h) {
      int mid = (l + h) / 2;
      if (k == a[mid]) return mid;
      if (k < a[mid]) {
        return binary_s(a, l, mid - 1, k);
      } else
        return binary_s(a, mid + 1, h, k);
    } else
      return -1;
  }
  void bubble_sort(char x[], int n) {
    for (int i = 0; i < n - 1; i++) {
      for (int j = 0; j < n - i - 1; j++) {
        if (x[j] > x[j + 1]) {
          char temp = x[j];
          x[j] = x[j + 1];
          x[j + 1] = temp;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      cout << x[i];
    }
    cout << '\n';
  }
  long long npr(int n, int r) {
    long long res = 1;
    for (int i = n - r + 1; i <= n; i++) {
      res *= i;
    }
    return res;
  }
  long long cper(int n, int r) {
    long long res = 1;
    for (int i = n - r + 1; i <= n; i++) {
      res *= i;
      if (res % r == 0) {
        res /= r;
        r = 1;
      }
    }
    return res;
  }
  long long ncr(int n, int r) { return npr(n, r) / factorial(r); }
  long long m;
  long long fast_power(long long base, long long power) {
    if (power == 0) {
      return 1;
    }
    long long res = fast_power(base, power / 2) % m;
    res *= res;
    res %= m;
    if (power % 2 == 0) {
      return res;
    } else {
      return (res * (base % m)) % m;
    }
  }
  long long sum(long long x) { return x * (x + 1) / 2; }
  long long gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long lcm(int a, int b) { return (a / gcd(a, b)) * b; }
  long long mod = 1e9 + 7;
  long long ipow(long long a, long long b) {
    long long answer = 1;
    for (; b--;) answer *= a;
    return answer;
  }
  int main() {
    speed();
    int t, n, arr[200005], arr1[31] = {0}, c;
    vector<int> ans;
    cin >> t;
    while (t--) {
      cin >> n;
      for (int i = 1; i <= n; i++) {
        cin >> arr[i];
      }
      for (int i = 1; i <= n; i++) {
        for (int j = 0; j < 31; j++) {
          if ((arr[i] >> j) & 1) arr1[j]++;
        }
      }
      for (int i = 1; i <= n; i++) {
        c = 0;
        for (int j = 0; j < 31; j++) {
          if (arr1[j] % i == 0) {
            c++;
          }
        }
        if (c == 31) {
          ans.push_back(i);
        }
      }
      for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
      for (int j = 0; j < 31; j++) {
        arr1[j] = 0;
      }
      ans.clear();
      cout << '\n';
    }
    return 0;
  }
  