It can be shown that any positive integer x can be uniquely represented as x = 1 + 2 + 4 + ... + 2k - 1 + r, where k and r are integers, k ≥ 0, 0 < r ≤ 2k. Let's call that representation prairie partition of x.

For example, the prairie partitions of 12, 17, 7 and 1 are: 

12 = 1 + 2 + 4 + 5,

17 = 1 + 2 + 4 + 8 + 2,

7 = 1 + 2 + 4,

1 = 1. 

Alice took a sequence of positive integers (possibly with repeating elements), replaced every element with the sequence of summands in its prairie partition, arranged the resulting numbers in non-decreasing order and gave them to Borys. Now Borys wonders how many elements Alice's original sequence could contain. Find all possible options!

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of numbers given from Alice to Borys.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1012; a1 ≤ a2 ≤ ... ≤ an) — the numbers given from Alice to Borys.

Output

Output, in increasing order, all possible values of m such that there exists a sequence of positive integers of length m such that if you replace every element with the summands in its prairie partition and arrange the resulting numbers in non-decreasing order, you will get the sequence given in the input.

If there are no such values of m, output a single integer -1.

Examples

Input

8
1 1 2 2 3 4 5 8


Output

2 


Input

6
1 1 1 2 2 2


Output

2 3 


Input

5
1 2 4 4 4


Output

-1

Note

In the first example, Alice could get the input sequence from [6, 20] as the original sequence.

In the second example, Alice's original sequence could be either [4, 5] or [3, 3, 3].
#include <bits/stdc++.h>
using namespace std;
struct cww {
  cww() {
    if (1) {
      ios::sync_with_stdio(false);
      cin.tie(0);
    }
  }
} star;
template <typename T>
inline bool chmin(T &l, T r) {
  bool a = l > r;
  if (a) l = r;
  return a;
}
template <typename T>
inline bool chmax(T &l, T r) {
  bool a = l < r;
  if (a) l = r;
  return a;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto &it : v) is >> it;
  return is;
}
int N;
long long a[112345];
vector<int> cnt(50, 0);
vector<long long> gomi;
int getmax() {
  auto rest = cnt;
  vector<long long> len;
  vector<long long> btk = gomi;
  while (rest[0] > 0) {
    for (int i = (0); i < (50); i++) {
      if (rest[i] == 0) {
        len.push_back(1ll << (i - 1));
        break;
      } else
        rest[i]--;
    }
  }
  for (int i = (0); i < (50); i++)
    while (rest[i]) {
      btk.push_back(1ll << i);
      rest[i]--;
    }
  sort((btk).begin(), (btk).end());
  reverse((btk).begin(), (btk).end());
  if (btk.size() > len.size()) return -1;
  for (int i = (0); i < (btk.size()); i++) {
    if (len[i] * 2 < btk[i]) return -1;
  }
  return len.size();
}
bool can(int sz) {
  auto rest = cnt;
  vector<long long> len;
  vector<long long> btk = gomi;
  for (int latte = (0); latte < (sz); latte++) {
    if (rest[0] == 0) return false;
    for (int i = (0); i < (50); i++) {
      if (rest[i] == 0) {
        len.push_back(1ll << (i - 1));
        break;
      } else
        rest[i]--;
    }
  }
  for (int i = (0); i < (50); i++)
    while (rest[i]) {
      btk.push_back(1ll << i);
      rest[i]--;
    }
  sort((btk).begin(), (btk).end());
  reverse((btk).begin(), (btk).end());
  if (btk.size() > len.size()) return false;
  for (int i = (0); i < (btk.size()); i++) {
    if (len[i] * 2 < btk[i]) return false;
  }
  return true;
}
int main() {
  cin >> N;
  for (int i = (0); i < (N); i++) cin >> a[i];
  for (int i = (0); i < (50); i++) {
    long long k = 1ll << i;
    for (int j = (0); j < (N); j++)
      if (a[j] == k) {
        a[j] = 0;
        cnt[i]++;
      }
  }
  for (int i = (0); i < (N); i++)
    if (a[i]) gomi.push_back(a[i]);
  int top = getmax();
  if (top == -1) {
    cout << -1 << endl;
    return 0;
  }
  int lb = 0;
  int ub = top;
  while (ub - lb > 1) {
    const int mid = (lb + ub) / 2;
    if (can(mid))
      ub = mid;
    else
      lb = mid;
  }
  int bottom = ub;
  for (int i = (bottom); i < (top); i++) cout << i << " ";
  cout << top << endl;
  return 0;
}
