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;
int cnt[60];
int ost[60];
vector<int> ans;
bool check(int m) {
  if (cnt[0] < m) return false;
  int now = m;
  int after[60];
  int newost[60];
  for (int i = 0; i < 60; i++) newost[i] = ost[i];
  for (int i = 0; i < 60; i++) {
    now = min(now, cnt[i]);
    after[i] = now;
    newost[i] += max(0, cnt[i] - now);
  }
  int k = 0;
  for (int i = 59; i >= 0; i--) {
    if (after[i] - k < newost[i]) return false;
    k += newost[i];
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long x;
    long long st = 1;
    int it = 0;
    cin >> x;
    while (st * 2 <= x) {
      st *= 2;
      it++;
    }
    if (x == st)
      cnt[it]++;
    else
      ost[it]++;
  }
  for (int m = 1; m <= n; m++) {
    if (check(m)) ans.push_back(m);
  }
  if (ans.empty())
    cout << -1 << endl;
  else
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << ' ';
}
