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:

$$$ r_i = \begin{cases} 1, ~ if ~ x_i = 1 ~ and ~ y_i = 1 \\\ 0, ~ if ~ x_i = 0 ~ or ~ y_i = 0 \end{cases} $$$
#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;
}
