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.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.OutputStream;
  import java.io.PrintWriter;
  import java.io.BufferedWriter;
  import java.util.InputMismatchException;
  import java.io.IOException;
  import java.util.ArrayList;
  import java.util.List;
  import java.io.Writer;
  import java.io.OutputStreamWriter;
  import java.io.InputStream;
  
  /**
   * Built using CHelper plug-in
   * Actual solution is at the top
   */
  public class Main {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          InputReader in = new InputReader(inputStream);
          OutputWriter out = new OutputWriter(outputStream);
          AArrayElimination solver = new AArrayElimination();
          int testCount = Integer.parseInt(in.next());
          for (int i = 1; i <= testCount; i++)
              solver.solve(i, in, out);
          out.close();
      }
  
      static class AArrayElimination {
          public void solve(int testNumber, InputReader in, OutputWriter out) {
              int n = in.nextInt();
              int[] arr = in.nextIntArray(n);
              int[] counts = new int[31];
              for (int i = 0; i < 32; i++) {
                  int mask = (1 << i);
                  for (int el : arr) {
                      if ((el & mask) > 0) {
                          counts[i]++;
                      }
                  }
              }
              List<Integer> res = new ArrayList<>();
              for (int i = 1; i <= n; i++) {
                  boolean divs = true;
                  for (int count : counts) {
                      if (count % i != 0) {
                          divs = false;
                          break;
                      }
                  }
                  if (divs)
                      res.add(i);
              }
              out.println(res.toArray());
          }
  
      }
  
      static class OutputWriter {
          private final PrintWriter writer;
  
          public OutputWriter(OutputStream outputStream) {
              writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
          }
  
          public OutputWriter(Writer writer) {
              this.writer = new PrintWriter(writer);
          }
  
          public void print(Object... objects) {
              print(" ", objects);
          }
  
          public void print(String sep, Object... objects) {
              for (int i = 0; i < objects.length; i++) {
                  if (i != 0) {
                      writer.print(sep);
                  }
                  writer.print(objects[i]);
              }
          }
  
          public void println(Object... objects) {
              print(objects);
              writer.println();
          }
  
          public void close() {
              writer.close();
          }
  
      }
  
      static class InputReader {
          final private int BUFFER_SIZE = 1 << 16;
          private InputStream stream;
          private byte[] buf = new byte[BUFFER_SIZE];
          private int curChar;
          private int numChars;
          private InputReader.SpaceCharFilter filter;
  
          public InputReader(InputStream stream) {
              this.stream = stream;
          }
  
          public int read() {
              if (numChars == -1) {
                  throw new InputMismatchException();
              }
              if (curChar >= numChars) {
                  curChar = 0;
                  try {
                      numChars = stream.read(buf);
                  } catch (IOException e) {
                      throw new InputMismatchException();
                  }
                  if (numChars <= 0) {
                      return -1;
                  }
              }
              return buf[curChar++];
          }
  
          public int nextInt() {
              int c = read();
              while (isSpaceChar(c)) {
                  c = read();
              }
              int sgn = 1;
              if (c == '-') {
                  sgn = -1;
                  c = read();
              }
              int res = 0;
              do {
                  if (c < '0' || c > '9') {
                      throw new InputMismatchException();
                  }
                  res *= 10;
                  res += c - '0';
                  c = read();
              } while (!isSpaceChar(c));
              return res * sgn;
          }
  
          public String nextString() {
              int c = read();
              while (isSpaceChar(c)) {
                  c = read();
              }
              StringBuilder res = new StringBuilder();
              do {
                  if (Character.isValidCodePoint(c)) {
                      res.appendCodePoint(c);
                  }
                  c = read();
              } while (!isSpaceChar(c));
              return res.toString();
          }
  
          public boolean isSpaceChar(int c) {
              if (filter != null) {
                  return filter.isSpaceChar(c);
              }
              return isWhitespace(c);
          }
  
          public static boolean isWhitespace(int c) {
              return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
          }
  
          public String next() {
              return nextString();
          }
  
          public int[] nextIntArray(int n) {
              int[] array = new int[n];
              for (int i = 0; i < n; ++i) array[i] = nextInt();
              return array;
          }
  
          public interface SpaceCharFilter {
              public boolean isSpaceChar(int ch);
  
          }
  
      }
  }
  
  