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: import java.io.*;
  import java.util.*;
  
  public class problem1601a {
  
      public static void main(String[] args){
  
          FastScanner sc = new FastScanner();
          int t = sc.nextInt();
          while(t-- > 0){
  
              int n = sc.nextInt();
              int arr[] = new int[n];
              for(int i=0; i<n; i++){
  
                  arr[i] = sc.nextInt();
  
              }
  
              int count[] = new int[32];
              for(int i=0; i<32; i++){
  
                  for(int j=0; j<n; j++){
  
                      if((arr[j] & 1) != 0){
                          count[i]++;
                      }
  
                      arr[j]/=2;
  
                  }
  
              }
  
              for(int i=1; i<=n; i++){
  
                  boolean doesDivide = true;
  
                  for(int j=0; j<32; j++){
  
                      if(count[j]%i != 0){
                          doesDivide = false;
                      }
  
                  }
  
                  if(doesDivide){
                      System.out.print(i + " ");
                  }
  
              }
  
              System.out.println();
  
          }
  
      }
  
  }
  
  class FastScanner
  {
      //I don't understand how this works lmao
      private int BS = 1 << 16;
      private char NC = (char) 0;
      private byte[] buf = new byte[BS];
      private int bId = 0, size = 0;
      private char c = NC;
      private double cnt = 1;
      private BufferedInputStream in;
  
      public FastScanner() {
          in = new BufferedInputStream(System.in, BS);
      }
  
      public FastScanner(String s) {
          try {
              in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
          } catch (Exception e) {
              in = new BufferedInputStream(System.in, BS);
          }
      }
  
      private char getChar() {
          while (bId == size) {
              try {
                  size = in.read(buf);
              } catch (Exception e) {
                  return NC;
              }
              if (size == -1) return NC;
              bId = 0;
          }
          return (char) buf[bId++];
      }
  
      public int nextInt() {
          return (int) nextLong();
      }
  
      public int[] nextInts(int N) {
          int[] res = new int[N];
          for (int i = 0; i < N; i++) {
              res[i] = (int) nextLong();
          }
          return res;
      }
  
      public long[] nextLongs(int N) {
          long[] res = new long[N];
          for (int i = 0; i < N; i++) {
              res[i] = nextLong();
          }
          return res;
      }
  
      public long nextLong() {
          cnt = 1;
          boolean neg = false;
          if (c == NC) c = getChar();
          for (; (c < '0' || c > '9'); c = getChar()) {
              if (c == '-') neg = true;
          }
          long res = 0;
          for (; c >= '0' && c <= '9'; c = getChar()) {
              res = (res << 3) + (res << 1) + c - '0';
              cnt *= 10;
          }
          return neg ? -res : res;
      }
  
      public double nextDouble() {
          double cur = nextLong();
          return c != '.' ? cur : cur + nextLong() / cnt;
      }
  
      public double[] nextDoubles(int N) {
          double[] res = new double[N];
          for (int i = 0; i < N; i++) {
              res[i] = nextDouble();
          }
          return res;
      }
  
      public String next() {
          StringBuilder res = new StringBuilder();
          while (c <= 32) c = getChar();
          while (c > 32) {
              res.append(c);
              c = getChar();
          }
          return res.toString();
      }
  
      public String nextLine() {
          StringBuilder res = new StringBuilder();
          while (c <= 32) c = getChar();
          while (c != '\n') {
              res.append(c);
              c = getChar();
          }
          return res.toString();
      }
  
      public boolean hasNext() {
          if (c > 32) return true;
          while (true) {
              c = getChar();
              if (c == NC) return false;
              else if (c > 32) return true;
          }
      }
  }