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 N = 1e5 + 5;
long long n, second, e, a[N], cnt[60];
vector<long long> vec(N, -1), rest;
int main() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (ceil(log2(a[i])) == floor(log2(a[i])))
      cnt[(long long)log2(a[i])]++;
    else
      rest.push_back(a[i]);
  }
  long long sz = 0, first;
  for (int i = 0; i < 60; i++) {
    int j = 0;
    while (cnt[i]) {
      if (vec[j] < i - 1) break;
      vec[j] = i, cnt[i]--, j++;
    }
  }
  for (long long i = 0; i < 60; i++)
    while (cnt[i]--) rest.push_back((1LL << i));
  sz = vec.size() - count(vec.begin(), vec.end(), -1);
  if (sz < rest.size()) return cout << -1 << endl, 0;
  sort(rest.rbegin(), rest.rend());
  for (long long k = 0; k < rest.size(); k++)
    if (rest[k] > (1LL << vec[k] + 1)) return cout << -1 << endl, 0;
  e = second = sz, sz -= rest.size();
  int i = 1e5 + 5;
  while (i--) {
    if (vec[i] == -1) continue;
    if (vec[i] + 1 < sz) sz -= (vec[i] + 2), second--;
  }
  for (; second <= e; second++) cout << second << " ";
  cout << endl;
}
