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 max_power = 50;
int two_powers[max_power];
int between[max_power];
int extra[max_power];
int ende[max_power];
int main() {
  int N;
  cin >> N;
  bool okay;
  int two_pow;
  long long tmp;
  for (int i = 0; i < N; i++) {
    cin >> tmp;
    two_pow = 0;
    okay = true;
    while (tmp > 1) {
      if (tmp & 1) okay = false;
      two_pow++;
      tmp /= 2;
    }
    if (okay)
      two_powers[two_pow]++;
    else
      between[two_pow]++;
  }
  int low = 1, high = two_powers[0] + 1, mid, cur;
  while (low < high) {
    mid = (low + high) / 2;
    cur = mid;
    for (int i = 0; i < max_power; i++) {
      extra[i] = two_powers[i] - cur;
      int nxt_cur = min(cur, two_powers[i + 1]);
      ende[i] = cur - nxt_cur;
      cur = nxt_cur;
    }
    bool fail = false;
    int open = 0;
    for (int i = max_power - 1; i >= 0; i--) {
      open += ende[i];
      open -= between[i];
      open -= extra[i + 1];
      if (open < 0) {
        fail = true;
        break;
      }
    }
    open -= extra[0];
    if (open < 0) fail = true;
    if (fail)
      low = mid + 1;
    else
      high = mid;
  }
  if (low > two_powers[0])
    cout << "-1";
  else
    for (int i = low; i <= two_powers[0]; i++) cout << i << " ";
  return 0;
}
