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:

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

#include <bits/stdc++.h>
using namespace std;

#if 1
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif

#if 0
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
template <typename T>
using extset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#endif

#if 0
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>
using Int = boost::multiprecision::cpp_int;
using lld = boost::multiprecision::cpp_dec_float_100;
#endif

#pragma region Macros

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
template <typename T>
using smaller_queue = priority_queue<T, vector<T>, greater<T>>;

constexpr ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353);
constexpr int INF = (int)1e9 + 10;
constexpr ll LINF = (ll)4e18;
constexpr ld PI = acos(-1.0);
constexpr ld EPS = 1e-7;
constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};
constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0};

#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define REPI(i, m, n) for (int i = m; i < (int)(n); ++i)
#define repi(i, n) REPI(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << "\n"
#define Yes(n) cout << ((n) ? "Yes" : "No") << "\n"
#define possible(n) cout << ((n) ? "possible" : "impossible") << "\n"
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << "\n"
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << "\n";
#define UNIQUE(v) v.erase(unique(all(v)), v.end())

struct io_init {
  io_init() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(30) << setiosflags(ios::fixed);
  };
} io_init;
template <typename T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
inline ll CEIL(ll a, ll b) { return (a + b - 1) / b; }
template <typename A, size_t N, typename T>
inline void Fill(A (&array)[N], const T &val) {
  fill((T *)array, (T *)(array + N), val);
}
template <typename T>
vector<T> compress(vector<T> &v) {
  vector<T> val = v;
  sort(all(val)), val.erase(unique(all(val)), val.end());
  for (auto &&vi : v) vi = lower_bound(all(val), vi) - val.begin();
  return val;
}
template <typename T, typename U>
constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept {
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
constexpr ostream &operator<<(ostream &os, pair<T, U> &p) noexcept {
  os << p.first << " " << p.second;
  return os;
}
random_device seed_gen;
mt19937_64 engine(seed_gen());

#pragma endregion

// --------------------------------------

void solve() {
  int n;
  cin >> n;
  vector<ll> a(n);
  rep(i, n) cin >> a[i];
  ll g = 0;
  rep(bit, 30) {
    ll cnt = 0;
    rep(i, n) {
      if (a[i] & (1 << i)) cnt++;
    }
    g = gcd(g, cnt);
  }
  vector<int> ans;
  if (g == 0) {
    REP(i, 1, n + 1) ans.push_back(i);
  } else {
    REP(i, 1, g + 1) {
      if (g % i == 0) ans.push_back(i);
    }
  }
  rep(i, ans.size()) cout << ans[i] << (i + 1 == ans.size() ? "\n" : " ");
}

int main() {
  int testcase;
  cin >> testcase;
  while (testcase--) solve();
}