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 = 200001;
int pw[63], cnt[63], n, pcnt[63], ppw[63];
long long a[MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    int j = 0;
    while (a[i] > (1LL << j)) {
      j++;
    }
    cnt[j]++;
    if (a[i] == (1LL << j)) {
      pw[j]++;
      mx = max(j, mx);
    }
  }
  vector<int> ans;
  for (int i = 1; i <= pw[0]; i++) {
    int last[63];
    memset(last, 0, sizeof(last));
    memcpy(ppw, pw, sizeof(pcnt));
    memcpy(pcnt, cnt, sizeof(pcnt));
    last[0] = i;
    ppw[0] -= i;
    pcnt[0] -= i;
    for (int j = 1; j <= mx; j++) {
      int mmn = min({last[j - 1], ppw[j]});
      last[j - 1] -= mmn;
      last[j] = mmn;
      ppw[j] -= mmn;
      pcnt[j] -= mmn;
    }
    for (int j = 0; j <= mx; j++) {
      for (int k = j + 1; k >= 0 && last[j]; k--) {
        int mmn = min(last[j], pcnt[k]);
        last[j] -= mmn;
        pcnt[k] -= mmn;
      }
    }
    bool ck = true;
    for (int i = 0; i < 63; i++) {
      if (pcnt[i]) ck = false;
    }
    if (ck) ans.push_back(i);
  }
  if (ans.size() == 0)
    cout << -1;
  else {
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i] << ' ';
    }
  }
  return 0;
}
