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;
multiset<long long> x;
multiset<long long> y;
int main() {
  int upper = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long tmp;
    cin >> tmp;
    if (tmp == 1) {
      upper++;
    }
    x.insert(tmp);
  }
  while (!x.empty()) {
    long long s = *x.begin();
    if (s != 1) {
      printf("-1\n");
      return 0;
    }
    while (true) {
      x.erase(x.find(s));
      s = s * 2;
      if (x.find(s) == x.end()) {
        y.insert(s / 2);
        break;
      }
    }
    if (y.size() >= x.size()) {
      break;
    }
  }
  for (auto i = y.rbegin(), j = x.rbegin(); j != x.rend(); i++, j++) {
    if (*j > *i * 2) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = y.size(); i <= upper; i++) {
    printf("%d ", i);
  }
  printf("\n");
}
