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;
const int MAXN = 1e5 + 15;
int n;
long long a[MAXN];
int pows[50], rest[50], lens[50];
int main() {
  cin.sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    int j;
    for (j = 0; (1LL << j) < a[i]; j++)
      ;
    if ((1LL << j) == a[i]) {
      pows[j]++;
    } else {
      rest[j]++;
    }
  }
  int base = pows[0];
  for (int i = 1; i < 50; i++) {
    if (pows[i] <= pows[i - 1]) {
      lens[i - 1] = pows[i - 1] - pows[i];
    } else {
      rest[i] += pows[i] - pows[i - 1];
      pows[i] = pows[i - 1];
    }
  }
  int len = 0;
  for (int i = 0; i < 50; i++) {
    len += lens[i];
  }
  for (int i = 45; i >= 0; i--) {
    if (rest[i] > 0) {
      for (int j = 45; j >= i - 1; j--) {
        if (lens[j] >= rest[i]) {
          lens[j] -= rest[i];
          rest[i] = 0;
        } else {
          rest[i] -= lens[j];
          lens[j] = 0;
        }
      }
      if (rest[i] > 0) {
        cout << -1;
        return 0;
      }
    }
  }
  int available = 0;
  for (int i = 0; i < 50; i++) {
    available += lens[i];
  }
  set<int> s;
  s.insert(len);
  int sum = 0, cnt = 0;
  for (int i = 0; i < 50; i++) {
    for (int j = 0; j < lens[i]; j++) {
      cnt++;
      sum += (i + 1);
      if (sum <= available - cnt) {
        s.insert(len - cnt);
      } else {
        break;
      }
    }
  }
  for (auto x : s) {
    cout << x << " ";
  }
  return 0;
}
