#include <bits/stdc++.h>
using namespace std;
namespace Eddd {
inline char sc() {
  return getchar();
  static const int LEN = 100000;
  static char Buf[LEN], *OP = Buf, *ED = Buf;
  if (OP == ED) ED = (OP = Buf) + fread(Buf, 1, LEN, stdin);
  return *OP++;
}
template <class T>
void rd(T &x) {
  static int c, f;
  x = f = 0;
  while (c = sc(), c < 48)
    if (c == '-') f = 1;
  do x = (x << 3) + (x << 1) + (c ^ 48);
  while (c = sc(), c > 47);
  if (f) x = -x;
}
template <class T>
void pt(T x) {
  if (x < 0)
    putchar('-'), x = -x;
  else if (!x)
    putchar('0');
  static int Stk[30], tp = 0;
  for (; x; x /= 10) Stk[tp++] = x % 10;
  while (tp) putchar(Stk[--tp] ^ 48);
}
template <class T>
void ptk(const T &x) {
  pt(x);
  putchar(' ');
}
template <class T>
void ptn(const T &x) {
  pt(x);
  putchar('\n');
}
template <class T>
bool chkmi(T &x, const T &y) {
  return x > y ? x = y, true : false;
}
template <class T>
bool chkmx(T &x, const T &y) {
  return x < y ? x = y, true : false;
}
template <class T>
T gcd(const T &x, const T &y) {
  return y ? gcd(y, x % y) : x;
}
}  // namespace Eddd
using namespace Eddd;
const int N = 500010;
const int Mod = 1e9 + 7;
long long Fast(long long x, long long b) {
  long long t = 1;
  for (; b; b >>= 1, x = x * x % Mod) {
    if (b & 1) t = t * x % Mod;
  }
  return t;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int a[N];
int ans[N];
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = (1), _t = (n); i <= _t; ++i) scanf("%d", &a[i]);
  int g = 0;
  for (int i = (0), _t = (29); i <= _t; ++i) {
    int c = 0;
    for (int j = (1), _t = (n); j <= _t; ++j)
      if ((a[j] >> i) & 1) ++c;
    g = gcd(c, g);
  }
  int m = 0;
  for (int i = (1), _t = (n); i <= _t; ++i)
    if (g % i == 0) ans[++m] = i;
  for (int i = (1), _t = (m); i <= _t; ++i)
    printf("%d%c", ans[i], " \n"[i == m]);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
using namespace std;
namespace Eddd {
inline char sc() {
  return getchar();
  static const int LEN = 100000;
  static char Buf[LEN], *OP = Buf, *ED = Buf;
  if (OP == ED) ED = (OP = Buf) + fread(Buf, 1, LEN, stdin);
  return *OP++;
}
template <class T>
void rd(T &x) {
  static int c, f;
  x = f = 0;
  while (c = sc(), c < 48)
    if (c == '-') f = 1;
  do x = (x << 3) + (x << 1) + (c ^ 48);
  while (c = sc(), c > 47);
  if (f) x = -x;
}
template <class T>
void pt(T x) {
  if (x < 0)
    putchar('-'), x = -x;
  else if (!x)
    putchar('0');
  static int Stk[30], tp = 0;
  for (; x; x /= 10) Stk[tp++] = x % 10;
  while (tp) putchar(Stk[--tp] ^ 48);
}
template <class T>
void ptk(const T &x) {
  pt(x);
  putchar(' ');
}
template <class T>
void ptn(const T &x) {
  pt(x);
  putchar('\n');
}
template <class T>
bool chkmi(T &x, const T &y) {
  return x > y ? x = y, true : false;
}
template <class T>
bool chkmx(T &x, const T &y) {
  return x < y ? x = y, true : false;
}
template <class T>
T gcd(const T &x, const T &y) {
  return y ? gcd(y, x % y) : x;
}
}  // namespace Eddd
using namespace Eddd;
const int N = 500010;
const int Mod = 1e9 + 7;
long long Fast(long long x, long long b) {
  long long t = 1;
  for (; b; b >>= 1, x = x * x % Mod) {
    if (b & 1) t = t * x % Mod;
  }
  return t;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int a[N];
int ans[N];
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = (1), _t = (n); i <= _t; ++i) scanf("%d", &a[i]);
  int g = 0;
  for (int i = (0), _t = (29); i <= _t; ++i) {
    int c = 0;
    for (int j = (1), _t = (n); j <= _t; ++j)
      if ((a[j] >> i) & 1) ++c;
    g = gcd(c, g);
  }
  int m = 0;
  for (int i = (1), _t = (n); i <= _t; ++i)
    if (g % i == 0) ans[++m] = i;
  for (int i = (1), _t = (m); i <= _t; ++i)
    printf("%d%c", ans[i], " \n"[i == m]);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

```