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 int n;
vector<long long int> powersOfTwo;
vector<long long int> residues;
vector<long long int> solutions;
long long int cp;
void process(long long int m) {
  long long int residuesAvailable = 0;
  long long int currentCatching = m;
  long long int lost;
  for (int i = 0; i <= cp + 1; i++) {
    if (powersOfTwo[i] >= currentCatching) {
      residuesAvailable += residues[i] + powersOfTwo[i] - currentCatching;
    } else if (i > 0) {
      lost = min(currentCatching - powersOfTwo[i], residuesAvailable);
      currentCatching = powersOfTwo[i];
      residuesAvailable += -lost + residues[i];
    } else {
      return;
    }
  }
  if (residuesAvailable == 0) {
    solutions.push_back(m);
  }
}
int main() {
  powersOfTwo = vector<long long int>(50, 0);
  residues = vector<long long int>(50, 0);
  solutions = vector<long long int>();
  long long int cpower = 1;
  cp = 0;
  long long int input;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> input;
    if (input == cpower) {
      powersOfTwo[cp]++;
    } else if (cpower < input && 2 * cpower > input) {
      residues[cp]++;
    } else {
      while (2 * cpower <= input) {
        cpower = cpower * 2;
        cp++;
      }
      if (input == cpower) {
        powersOfTwo[cp]++;
      } else {
        residues[cp]++;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    process(i);
  }
  if (solutions.size() == 0) {
    cout << -1 << endl;
  } else {
    for (int i = 0; i < solutions.size(); i++) {
      cout << solutions[i] << " ";
    }
    cout << endl;
  }
}
