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;
bool cmp(int a, int b) { return a > b; }
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
  return a * (b / gcd(a, b));
}
long long n, a[5000005], pw[300], btw[300];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    for (long long j = 0; j < 50; j++) {
      long long val = 1LL << j;
      if (a[i] < val) {
        btw[j - 1]++;
        break;
      } else if (a[i] == val) {
        pw[j]++;
        break;
      }
    }
  }
  bool fd = false;
  for (long long i = 1; i <= pw[0]; i++) {
    long long ct = 0;
    long long need = i;
    for (long long j = 0; j < 50; j++) {
      if (pw[j] >= need) {
        ct += pw[j] - need;
      } else {
        ct -= need - pw[j];
        ct = max(0LL, ct);
        need = pw[j];
      }
      ct += btw[j];
    }
    if (ct == 0) {
      cout << i << " ";
      fd = true;
    }
  }
  if (!fd) {
    cout << -1;
  }
  cout << endl;
  return 0;
}
