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;
using ll = long long;
using ld = double;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<ld, ld>;
using cd = complex<ld>;
using vcd = vector<cd>;
using vi = vector<int>;
using vl = vector<ll>;
using vd = vector<ld>;
using vs = vector<string>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
template <class T>
using min_pq = priority_queue<T, vector<T>, greater<T> >;
template <class T>
inline int ckmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int ckmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
template <class T>
void remDup(vector<T>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), end(v));
}
constexpr int pct(int x) { return __builtin_popcount(x); }
constexpr int bits(int x) { return x == 0 ? 0 : 31 - __builtin_clz(x); }
constexpr int p2(int x) { return 1 << x; }
constexpr int msk2(int x) { return p2(x) - 1; }
ll ceilDiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
ll floorDiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
void setPrec(int x) { cout << fixed << setprecision(x); }
string to_string(char c) { return string(1, c); }
string to_string(const char* s) { return (string)s; }
string to_string(string s) { return s; }
string to_string(bool b) { return to_string((int)b); }
template <class T>
string to_string(complex<T> c) {
  stringstream ss;
  ss << c;
  return ss.str();
}
template <class T>
using V = vector<T>;
string to_string(V<bool> v) {
  string res = "{";
  for (int i = (0); i <= (int((v).size()) - 1); i++) res += char('0' + v[i]);
  res += "}";
  return res;
}
template <size_t sz>
string to_string(bitset<sz> b) {
  string res = "";
  for (int i = (0); i <= (int((b).size()) - 1); i++) res += char('0' + b[i]);
  return res;
}
template <class T, class U>
string to_string(pair<T, U> p);
template <class T>
string to_string(T v) {
  bool fst = 1;
  string res = "";
  for (const auto& x : v) {
    if (!fst) res += " ";
    fst = 0;
    res += to_string(x);
  }
  return res;
}
template <class T, class U>
string to_string(pair<T, U> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <class T>
void pr(T x) {
  cout << to_string(x);
}
template <class T, class... U>
void pr(const T& t, const U&... u) {
  pr(t);
  pr(u...);
}
void ps() { pr("\n"); }
template <class T, class... U>
void ps(const T& t, const U&... u) {
  pr(t);
  if (sizeof...(u)) pr(" ");
  ps(u...);
}
void DBG() { cerr << "]" << endl; }
template <class T, class... U>
void DBG(const T& t, const U&... u) {
  cerr << to_string(t);
  if (sizeof...(u)) cerr << ", ";
  DBG(u...);
}
const ld PI = acos(-1.0);
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const ld EPS = 1e-9;
const ll MODBASE = 1000000007LL;
const int INF = 0x3f3f3f3f;
const int MAXN = 110;
const int MAXM = 1000;
const int MAXK = 16;
const int MAXQ = 200010;
int f[30], n;
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  int te;
  cin >> te;
  while (te--) {
    cin >> n;
    memset(f, 0, sizeof(f));
    for (int i = (1); i <= (n); i++) {
      int x;
      cin >> x;
      for (int j = (29); j >= (0); j--)
        if (x & (1 << j)) f[j]++;
    }
    int g = 0;
    for (int i = (0); i <= (29); i++) g = gcd(g, f[i]);
    if (g == 0) {
      for (int i = (1); i <= (n); i++) cout << i << " ";
      cout << "\n";
    } else {
      for (int i = (1); i <= (g); i++)
        if (n % i == 0) cout << i << ' ';
      cout << "\n";
    }
  }
  return 0;
}
