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, sum, freq[50];
vector<int> ans, v;
vector<long long> temp;
void no() {
  puts("-1");
  exit(0);
}
int main() {
  scanf("%d", &n);
  long long a;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a);
    if (a & (a - 1)) {
      temp.push_back(a);
      continue;
    }
    int count = 0;
    while (a > 1) {
      count++;
      a >>= 1;
    }
    freq[count]++;
  }
  sum = n - freq[0];
  v.resize(freq[0]);
  for (long long i = 1; i < 50; i++) {
    for (int j = 0; j < v.size() && freq[i]; j++, freq[i]--) {
      if (v[j] + 1 != i) break;
      sum--;
      v[j]++;
    }
  }
  for (int i = 1; i < 50; i++)
    while (freq[i]--) temp.push_back(1LL << i);
  sort(temp.rbegin(), temp.rend());
  if (v.size() < sum) no();
  for (int i = 0; i < temp.size(); i++)
    if (temp[i] > (1LL << (v[i] + 1))) no();
  ans.push_back(v.size());
  while (v.size() > 1) {
    if (v.size() - 1 < sum + v.back() + 1) break;
    for (int i = 0; i < v.back() + 1; i++)
      if ((1LL << (v.back() - i)) > (1LL << (v[sum + i] + 1))) no();
    sum += v.back() + 1;
    v.pop_back();
    ans.push_back(v.size());
  }
  reverse(ans.begin(), ans.end());
  for (auto i : ans) printf("%d ", i);
}
