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 main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> cnt(100);
  vector<int> cnt1(100);
  int maxe = 0;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    int e = 0;
    long long p2 = 1;
    while (p2 < x) {
      p2 *= 2LL;
      e++;
    }
    maxe = max(maxe, e);
    if (p2 == x) {
      cnt[e]++;
    } else {
      cnt1[e]++;
    }
  }
  vector<int> ans;
  for (int c = 1; c <= cnt[0]; c++) {
    int cc = c;
    int nu = cnt[0] - c;
    bool ok = true;
    for (int i = 1; i <= maxe; i++) {
      nu += cnt1[i];
      if (cnt[i] >= cc) {
        nu += cnt[i] - cc;
      } else if (nu < cc - cnt[i]) {
        nu = 0;
        cc = cnt[i];
      } else {
        nu -= (cc - cnt[i]);
        cc = cnt[i];
      }
    }
    if (nu > cc) {
      ok = false;
    }
    if (ok) {
      ans.push_back(c);
    }
  }
  if (!ans.size()) {
    cout << "-1\n";
    return 0;
  }
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i];
    if (i + 1 < ans.size()) {
      cout << " ";
    }
  }
  cout << "\n";
}
