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;
int n;
vector<long long> a;
bool check(int mid) {
  multiset<long long> s(a.begin(), a.end());
  vector<long long> c;
  for (int i = 0; i < mid; i++) {
    long long j = 1;
    while (s.find(j) != s.end()) {
      s.erase(s.find(j));
      j *= 2;
    }
    j /= 2;
    c.push_back(j);
  }
  reverse(c.begin(), c.end());
  while (s.size()) {
    long long cur = *--s.end();
    if (c.size() == 0) return 0;
    if (cur > c.back() * 2) return 0;
    s.erase(--s.end());
    c.pop_back();
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  int ones = 0;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    a.push_back(x);
    ones += x == 1;
  }
  int l = 1, r = ones;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (check(mid))
      r = mid - 1;
    else
      l = mid + 1;
  }
  if (r + 1 > ones) return cout << -1 << endl, 0;
  for (int i = r + 1; i <= ones; i++) cout << i << " ";
  cout << endl;
}
