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.*;
  
  import static java.lang.Math.*;
  import static java.util.Arrays.*;
  
  public class cf1601a {
  
      public static void main(String[] args) throws IOException {
          int t = ri();
          next: while (t --> 0) {
              int n = ri(), a[] = ria(n), cnt[] = new int[30];
              for (int i = 0; i < n; ++i) {
                  for (int j = 0; j < 30; ++j) {
                      if ((a[i] & (1 << j)) != 0) {
                          ++cnt[j];
                      }
                  }
              }
              List<Integer> ans = new ArrayList<>();
              ans.add(1);
              nxt: for (int i = 2; i <= n; ++i) {
                  for (int j = 0; j < 30; ++j) {
                      if (cnt[j] % i != 0) {
                          continue nxt;
                      }
                  }
                  ans.add(i);
              }
              prln(ans);
          }
          close();
      }
  
      static BufferedReader __i = new BufferedReader(new InputStreamReader(System.in));
      static PrintWriter __o = new PrintWriter(new OutputStreamWriter(System.out));
      static StringTokenizer input;
      static Random __r = new Random();
  
      // references
      // IBIG = 1e9 + 7
      // IMAX ~= 2e9
      // LMAX ~= 9e18
  
      // constants
      static final int IBIG = 1000000007;
      static final int IMAX = 2147483647;
      static final long LMAX = 9223372036854775807L;
      // math util
      static int minof(int a, int b, int c) {return min(a, min(b, c));}
      static int minof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); int min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
      static long minof(long a, long b, long c) {return min(a, min(b, c));}
      static long minof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); long min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
      static int maxof(int a, int b, int c) {return max(a, max(b, c));}
      static int maxof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); int max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
      static long maxof(long a, long b, long c) {return max(a, max(b, c));}
      static long maxof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); long max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
      static int powi(int a, int b) {if (a == 0) return 0; int ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
      static long powl(long a, int b) {if (a == 0) return 0; long ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
      static int fli(double d) {return (int) d;}
      static int cei(double d) {return (int) ceil(d);}
      static long fll(double d) {return (long) d;}
      static long cel(double d) {return (long) ceil(d);}
      static int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}
      static long gcd(long a, long b) {return b == 0 ? a : gcd(b, a % b);}
      static int[] exgcd(int a, int b) {if (b == 0) return new int[] {1, 0}; int[] y = exgcd(b, a % b); return new int[] {y[1], y[0] - y[1] * (a / b)};}
      static long[] exgcd(long a, long b) {if (b == 0) return new long[] {1, 0}; long[] y = exgcd(b, a % b); return new long[] {y[1], y[0] - y[1] * (a / b)};}
      static int randInt(int min, int max) {return __r.nextInt(max - min + 1) + min;}
      static long mix(long x) {x += 0x9e3779b97f4a7c15L; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L; x = (x ^ (x >> 27)) * 0x94d049bb133111ebL; return x ^ (x >> 31);}
      // array util
      static void reverse(int[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {int swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
      static void reverse(long[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {long swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
      static void reverse(double[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {double swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
      static void reverse(char[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {char swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
      static void shuffle(int[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
      static void shuffle(long[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
      static void shuffle(double[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
      static void rsort(int[] a) {shuffle(a); sort(a);}
      static void rsort(long[] a) {shuffle(a); sort(a);}
      static void rsort(double[] a) {shuffle(a); sort(a);}
      static int[] copy(int[] a) {int[] ans = new int[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
      static long[] copy(long[] a) {long[] ans = new long[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
      static double[] copy(double[] a) {double[] ans = new double[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
      static char[] copy(char[] a) {char[] ans = new char[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
      // input
      static void r() throws IOException {input = new StringTokenizer(rline());}
      static int ri() throws IOException {return Integer.parseInt(rline());}
      static long rl() throws IOException {return Long.parseLong(rline());}
      static double rd() throws IOException {return Double.parseDouble(rline());}
      static int[] ria(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni(); return a;}
      static void ria(int[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = ni();}
      static int[] riam1(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1; return a;}
      static void riam1(int[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1;}
      static long[] rla(int n) throws IOException {long[] a = new long[n]; r(); for (int i = 0; i < n; ++i) a[i] = nl(); return a;}
      static void rla(long[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = nl();}
      static double[] rda(int n) throws IOException {double[] a = new double[n]; r(); for (int i = 0; i < n; ++i) a[i] = nd(); return a;}
      static void rda(double[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = nd();}
      static char[] rcha() throws IOException {return rline().toCharArray();}
      static void rcha(char[] a) throws IOException {int n = a.length, i = 0; for (char c : rline().toCharArray()) a[i++] = c;}
      static String rline() throws IOException {return __i.readLine();}
      static String n() {return input.nextToken();}
      static int rni() throws IOException {r(); return ni();}
      static int ni() {return Integer.parseInt(n());}
      static long rnl() throws IOException {r(); return nl();}
      static long nl() {return Long.parseLong(n());}
      static double rnd() throws IOException {r(); return nd();}
      static double nd() {return Double.parseDouble(n());}
      // output
      static void pr(int i) {__o.print(i);}
      static void prln(int i) {__o.println(i);}
      static void pr(long l) {__o.print(l);}
      static void prln(long l) {__o.println(l);}
      static void pr(double d) {__o.print(d);}
      static void prln(double d) {__o.println(d);}
      static void pr(char c) {__o.print(c);}
      static void prln(char c) {__o.println(c);}
      static void pr(char[] s) {__o.print(new String(s));}
      static void prln(char[] s) {__o.println(new String(s));}
      static void pr(String s) {__o.print(s);}
      static void prln(String s) {__o.println(s);}
      static void pr(Object o) {__o.print(o);}
      static void prln(Object o) {__o.println(o);}
      static void prln() {__o.println();}
      static void pryes() {prln("yes");}
      static void pry() {prln("Yes");}
      static void prY() {prln("YES");}
      static void prno() {prln("no");}
      static void prn() {prln("No");}
      static void prN() {prln("NO");}
      static boolean pryesno(boolean b) {prln(b ? "yes" : "no"); return b;};
      static boolean pryn(boolean b) {prln(b ? "Yes" : "No"); return b;}
      static boolean prYN(boolean b) {prln(b ? "YES" : "NO"); return b;}
      static void prln(int... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
      static void prln(long... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
      static void prln(double... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
      static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for (int i = 0; i < n; pr(iter.next()), pr(' '), ++i); if (n >= 0) prln(iter.next()); else prln();}
      static void h() {prln("hlfd"); flush();}
      static void flush() {__o.flush();}
      static void close() {__o.close();}
  }