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 = 100010;
long long a[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n;
  cin >> n;
  int ma = 0;
  for (int j = 1; j <= n; ++j) {
    cin >> a[j];
    if (a[j] == 1) ++ma;
  }
  int r = ma + 1, l = 0;
  while (l + 1 < r) {
    int mid = (l + r) / 2;
    int mi = mid;
    int po = 1;
    queue<long long> oths;
    for (long long num = 1; num <= 1e13; num = num * 2) {
      int cs = 0;
      while (po <= n && a[po] <= num) {
        if (a[po] == num && cs < mi)
          ++cs;
        else
          oths.push(a[po]);
        ++po;
      }
      if (cs < mi) {
        int cha = mi - cs;
        for (int j = 1; j <= cha; ++j)
          if (oths.empty() == 0 && num >= oths.front()) oths.pop();
        mi = cs;
      }
    }
    if (oths.empty())
      r = mid;
    else
      l = mid;
  }
  if (r == ma + 1)
    cout << "-1" << endl;
  else {
    for (int j = r; j <= ma; ++j) cout << j << " ";
    cout << endl;
  }
  return 0;
}
