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;
long long n, a[100010], powNum[43];
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    for (long long j = 0; j < 43; ++j) {
      if (a[i] == (1LL << j)) {
        ++powNum[j];
      }
    }
  }
  bool ok = 0;
  for (long long i = 0; i <= n; ++i) {
    long long used = 0, curTops = i;
    if (powNum[0] < i) {
      break;
    }
    for (long long j = 0; j < 43; ++j) {
      used += min(powNum[j], curTops);
      long long freeSpots = max(0LL, curTops - powNum[j]);
      long long toAdd = (upper_bound(a, a + n, (1LL << j)) - a) - used;
      used += min(freeSpots, toAdd);
      curTops = min(curTops, powNum[j]);
      if (!curTops) {
        break;
      }
    }
    if (used == n) {
      ok = 1;
      cout << i << ' ';
    }
  }
  if (!ok) {
    cout << -1;
  }
}
