#include <bits/stdc++.h>
namespace fastIO {
const int maxc = 1 << 23;
char ibuf[maxc], *__p1 = ibuf, *__p2 = ibuf;
inline char getchar() {
  return __p1 == __p2 && (__p2 = (__p1 = ibuf) + fread(ibuf, 1, maxc, stdin),
                          __p1 == __p2)
             ? EOF
             : *__p1++;
}
template <typename T>
void read(T &n) {
  int x = 0;
  n = 0;
  char c = getchar();
  while (!isdigit(c)) x |= (c == '-'), c = getchar();
  while (isdigit(c)) n = n * 10 + c - '0', c = getchar();
  n = x ? -n : n;
}
void read(char *s) {
  int p = 0;
  char c = getchar();
  while (!isdigit(c) && !isalpha(c)) c = getchar();
  while (isalpha(c) || isdigit(c)) s[p++] = c, c = getchar();
  return;
}
char obuf[maxc], *__pO = obuf;
inline void putchar(char c) { *__pO++ = c; }
template <typename T>
void print(const T &x) {
  if (x < 0)
    putchar('-'), print(-x);
  else {
    if (x > 9) print(x / 10);
    putchar(x % 10 + '0');
  }
  return;
}
template <typename T>
inline void print(const T &x, const char &c) {
  print(x), putchar(c);
}
void output() { fwrite(obuf, __pO - obuf, 1, stdout); }
}  // namespace fastIO
using namespace fastIO;
template <typename T>
inline T max(const T &a, const T &b) {
  return a > b ? a : b;
}
template <typename T>
inline T min(const T &a, const T &b) {
  return a < b ? a : b;
}
template <typename T>
inline T chkmax(T &a, const T &b) {
  return a = max(a, b);
}
template <typename T>
inline T chkmin(T &a, const T &b) {
  return a = min(a, b);
}
template <typename T>
inline T myabs(const T &a) {
  return a >= 0 ? a : -a;
}
template <typename T>
inline void myswap(T &a, T &b) {
  T t = a;
  a = b, b = t;
  return;
}
const int maxn = 2e5 + 5;
int a[maxn], n;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main() {
  int T;
  read(T);
  while (T--) {
    read(n);
    bool flag = 1;
    for (int i = (1); i <= (n); ++i) read(a[i]), flag &= (a[i] == 0);
    if (flag) {
      for (int i = (1); i <= (n); ++i) print(i, ' ');
      putchar('\n');
      continue;
    }
    int g = 0;
    for (int j = (0); j <= (29); ++j) {
      int cnt = 0;
      for (int i = (1); i <= (n); ++i)
        if (a[i] & (1 << j)) ++cnt;
      g = gcd(cnt, g);
    }
    for (int i = (1); i <= (n); ++i)
      if (g % i == 0) print(i, ' ');
    putchar('\n');
  }
  return output(), 0;
}

### Prompt

Please formulate 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>
namespace fastIO {
const int maxc = 1 << 23;
char ibuf[maxc], *__p1 = ibuf, *__p2 = ibuf;
inline char getchar() {
  return __p1 == __p2 && (__p2 = (__p1 = ibuf) + fread(ibuf, 1, maxc, stdin),
                          __p1 == __p2)
             ? EOF
             : *__p1++;
}
template <typename T>
void read(T &n) {
  int x = 0;
  n = 0;
  char c = getchar();
  while (!isdigit(c)) x |= (c == '-'), c = getchar();
  while (isdigit(c)) n = n * 10 + c - '0', c = getchar();
  n = x ? -n : n;
}
void read(char *s) {
  int p = 0;
  char c = getchar();
  while (!isdigit(c) && !isalpha(c)) c = getchar();
  while (isalpha(c) || isdigit(c)) s[p++] = c, c = getchar();
  return;
}
char obuf[maxc], *__pO = obuf;
inline void putchar(char c) { *__pO++ = c; }
template <typename T>
void print(const T &x) {
  if (x < 0)
    putchar('-'), print(-x);
  else {
    if (x > 9) print(x / 10);
    putchar(x % 10 + '0');
  }
  return;
}
template <typename T>
inline void print(const T &x, const char &c) {
  print(x), putchar(c);
}
void output() { fwrite(obuf, __pO - obuf, 1, stdout); }
}  // namespace fastIO
using namespace fastIO;
template <typename T>
inline T max(const T &a, const T &b) {
  return a > b ? a : b;
}
template <typename T>
inline T min(const T &a, const T &b) {
  return a < b ? a : b;
}
template <typename T>
inline T chkmax(T &a, const T &b) {
  return a = max(a, b);
}
template <typename T>
inline T chkmin(T &a, const T &b) {
  return a = min(a, b);
}
template <typename T>
inline T myabs(const T &a) {
  return a >= 0 ? a : -a;
}
template <typename T>
inline void myswap(T &a, T &b) {
  T t = a;
  a = b, b = t;
  return;
}
const int maxn = 2e5 + 5;
int a[maxn], n;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main() {
  int T;
  read(T);
  while (T--) {
    read(n);
    bool flag = 1;
    for (int i = (1); i <= (n); ++i) read(a[i]), flag &= (a[i] == 0);
    if (flag) {
      for (int i = (1); i <= (n); ++i) print(i, ' ');
      putchar('\n');
      continue;
    }
    int g = 0;
    for (int j = (0); j <= (29); ++j) {
      int cnt = 0;
      for (int i = (1); i <= (n); ++i)
        if (a[i] & (1 << j)) ++cnt;
      g = gcd(cnt, g);
    }
    for (int i = (1); i <= (n); ++i)
      if (g % i == 0) print(i, ' ');
    putchar('\n');
  }
  return output(), 0;
}

```