#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
using namespace std;
template <class T>
using rque = priority_queue<T, vector<T>, greater<T>>;
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmax(T &a, const T &b) {
  if (b > a) {
    a = b;
    return 1;
  }
  return 0;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  long long cnt = a % b;
  while (cnt != 0) {
    a = b;
    b = cnt;
    cnt = a % b;
  }
  return b;
}
long long extGCD(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extGCD(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
struct UnionFind {
  vector<long long> data;
  int num;
  UnionFind(int sz) {
    data.assign(sz, -1);
    num = sz;
  }
  bool unite(int x, int y) {
    x = find(x), y = find(y);
    if (x == y) return (false);
    if (data[x] > data[y]) swap(x, y);
    data[x] += data[y];
    data[y] = x;
    num--;
    return (true);
  }
  int find(int k) {
    if (data[k] < 0) return (k);
    return (data[k] = find(data[k]));
  }
  long long size(int k) { return (-data[find(k)]); }
  bool same(int x, int y) { return find(x) == find(y); }
};
template <int mod>
struct ModInt {
  int x;
  ModInt() : x(0) {}
  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
  ModInt &operator+=(const ModInt &p) {
    if ((x += p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator-=(const ModInt &p) {
    if ((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator*=(const ModInt &p) {
    x = (int)(1LL * x * p.x % mod);
    return *this;
  }
  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }
  ModInt operator-() const { return ModInt(-x); }
  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
  ModInt &operator++() { return *this += ModInt(1); }
  ModInt operator++(int) {
    ModInt tmp = *this;
    ++*this;
    return tmp;
  }
  ModInt &operator--() { return *this -= ModInt(1); }
  ModInt operator--(int) {
    ModInt tmp = *this;
    --*this;
    return tmp;
  }
  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
  bool operator==(const ModInt &p) const { return x == p.x; }
  bool operator!=(const ModInt &p) const { return x != p.x; }
  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while (b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }
  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while (n > 0) {
      if (n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }
  friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x; }
  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt<mod>(t);
    return (is);
  }
  static int get_mod() { return mod; }
};
long long mpow2(long long x, long long n, long long mod) {
  long long ans = 1;
  while (n != 0) {
    if (n & 1) ans = ans * x % mod;
    x = x * x % mod;
    n = n >> 1;
  }
  return ans;
}
long long modinv2(long long a, long long mod) {
  long long b = mod, u = 1, v = 0;
  while (b) {
    long long t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= mod;
  if (u < 0) u += mod;
  return u;
}
constexpr int mod = 998244353;
using mint = ModInt<mod>;
mint mpow(mint x, long long n) {
  mint ans = 1;
  while (n != 0) {
    if (n & 1) ans *= x;
    x *= x;
    n = n >> 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  cout << fixed << setprecision(15);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (long long i = 0; i < (n); i++) cin >> a[i];
    int g = 0;
    for (long long t = 0; t < (30); t++) {
      int cnt = 0;
      for (long long i = 0; i < (n); i++) cnt += (a[i] >> t & 1);
      g = gcd(g, cnt);
    }
    vector<int> ans;
    for (long long i = 1; i < (n + 1); i++)
      if (gcd(g, i) == i) ans.emplace_back(i);
    for (long long i = 0; i < (ans.size()); i++)
      cout << ans[i] << (i + 1 == ans.size() ? '\n' : ' ');
  }
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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:

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

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
using namespace std;
template <class T>
using rque = priority_queue<T, vector<T>, greater<T>>;
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmax(T &a, const T &b) {
  if (b > a) {
    a = b;
    return 1;
  }
  return 0;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  long long cnt = a % b;
  while (cnt != 0) {
    a = b;
    b = cnt;
    cnt = a % b;
  }
  return b;
}
long long extGCD(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extGCD(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
struct UnionFind {
  vector<long long> data;
  int num;
  UnionFind(int sz) {
    data.assign(sz, -1);
    num = sz;
  }
  bool unite(int x, int y) {
    x = find(x), y = find(y);
    if (x == y) return (false);
    if (data[x] > data[y]) swap(x, y);
    data[x] += data[y];
    data[y] = x;
    num--;
    return (true);
  }
  int find(int k) {
    if (data[k] < 0) return (k);
    return (data[k] = find(data[k]));
  }
  long long size(int k) { return (-data[find(k)]); }
  bool same(int x, int y) { return find(x) == find(y); }
};
template <int mod>
struct ModInt {
  int x;
  ModInt() : x(0) {}
  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
  ModInt &operator+=(const ModInt &p) {
    if ((x += p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator-=(const ModInt &p) {
    if ((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator*=(const ModInt &p) {
    x = (int)(1LL * x * p.x % mod);
    return *this;
  }
  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }
  ModInt operator-() const { return ModInt(-x); }
  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
  ModInt &operator++() { return *this += ModInt(1); }
  ModInt operator++(int) {
    ModInt tmp = *this;
    ++*this;
    return tmp;
  }
  ModInt &operator--() { return *this -= ModInt(1); }
  ModInt operator--(int) {
    ModInt tmp = *this;
    --*this;
    return tmp;
  }
  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
  bool operator==(const ModInt &p) const { return x == p.x; }
  bool operator!=(const ModInt &p) const { return x != p.x; }
  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while (b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }
  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while (n > 0) {
      if (n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }
  friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x; }
  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt<mod>(t);
    return (is);
  }
  static int get_mod() { return mod; }
};
long long mpow2(long long x, long long n, long long mod) {
  long long ans = 1;
  while (n != 0) {
    if (n & 1) ans = ans * x % mod;
    x = x * x % mod;
    n = n >> 1;
  }
  return ans;
}
long long modinv2(long long a, long long mod) {
  long long b = mod, u = 1, v = 0;
  while (b) {
    long long t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= mod;
  if (u < 0) u += mod;
  return u;
}
constexpr int mod = 998244353;
using mint = ModInt<mod>;
mint mpow(mint x, long long n) {
  mint ans = 1;
  while (n != 0) {
    if (n & 1) ans *= x;
    x *= x;
    n = n >> 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  cout << fixed << setprecision(15);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (long long i = 0; i < (n); i++) cin >> a[i];
    int g = 0;
    for (long long t = 0; t < (30); t++) {
      int cnt = 0;
      for (long long i = 0; i < (n); i++) cnt += (a[i] >> t & 1);
      g = gcd(g, cnt);
    }
    vector<int> ans;
    for (long long i = 1; i < (n + 1); i++)
      if (gcd(g, i) == i) ans.emplace_back(i);
    for (long long i = 0; i < (ans.size()); i++)
      cout << ans[i] << (i + 1 == ans.size() ? '\n' : ' ');
  }
}

```