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>
  namespace fastIO {
  const int maxc = 1 << 23;
  char ibuf[maxc], *__p1 = ibuf, *__p2 = ibuf;
  inline char getchar() {
    return __p1 == __p2 && (__p2 = (__p1 = ibuf) + fread(ibuf, 1, maxc, stdin),
                            __p1 == __p2)
               ? EOF
               : *__p1++;
  }
  template <typename T>
  void read(T &n) {
    int x = 0;
    n = 0;
    char c = getchar();
    while (!isdigit(c)) x |= (c == '-'), c = getchar();
    while (isdigit(c)) n = n * 10 + c - '0', c = getchar();
    n = x ? -n : n;
  }
  void read(char *s) {
    int p = 0;
    char c = getchar();
    while (!isdigit(c) && !isalpha(c)) c = getchar();
    while (isalpha(c) || isdigit(c)) s[p++] = c, c = getchar();
    return;
  }
  char obuf[maxc], *__pO = obuf;
  inline void putchar(char c) { *__pO++ = c; }
  template <typename T>
  void print(const T &x) {
    if (x < 0)
      putchar('-'), print(-x);
    else {
      if (x > 9) print(x / 10);
      putchar(x % 10 + '0');
    }
    return;
  }
  template <typename T>
  inline void print(const T &x, const char &c) {
    print(x), putchar(c);
  }
  void output() { fwrite(obuf, __pO - obuf, 1, stdout); }
  }  // namespace fastIO
  using namespace fastIO;
  template <typename T>
  inline T max(const T &a, const T &b) {
    return a > b ? a : b;
  }
  template <typename T>
  inline T min(const T &a, const T &b) {
    return a < b ? a : b;
  }
  template <typename T>
  inline T chkmax(T &a, const T &b) {
    return a = max(a, b);
  }
  template <typename T>
  inline T chkmin(T &a, const T &b) {
    return a = min(a, b);
  }
  template <typename T>
  inline T myabs(const T &a) {
    return a >= 0 ? a : -a;
  }
  template <typename T>
  inline void myswap(T &a, T &b) {
    T t = a;
    a = b, b = t;
    return;
  }
  const int maxn = 2e5 + 5;
  int a[maxn], n;
  int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
  int main() {
    int T;
    read(T);
    while (T--) {
      read(n);
      bool flag = 1;
      for (int i = (1); i <= (n); ++i) read(a[i]), flag &= (a[i] == 0);
      if (flag) {
        for (int i = (1); i <= (n); ++i) print(i, ' ');
        putchar('\n');
        continue;
      }
      int g = 0;
      for (int j = (0); j <= (29); ++j) {
        int cnt = 0;
        for (int i = (1); i <= (n); ++i)
          if (a[i] & (1 << j)) ++cnt;
        g = gcd(cnt, g);
      }
      for (int i = (1); i <= (n); ++i)
        if (g % i == 0) print(i, ' ');
      putchar('\n');
    }
    return output(), 0;
  }
  