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 N = 1e6 + 1;
const int M = 1e9 + 7;
const double eps = 1e-6;
const double PI(acos(-1.0));
int n;
int p[N], b[N];
int f(long long m) {
  long long lef = p[0] - m;
  for (int i = (1); i <= (40); i++) {
    if (m >= p[i]) {
      lef -= min(lef, m - p[i]);
      m = p[i];
    } else
      lef += p[i] - m;
    lef += b[i];
  }
  return lef == 0;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  long long x;
  for (int i = 0; i < (n); i++) {
    cin >> x;
    int j = 0, f = 0;
    while (x > 1) {
      if (x & 1) f = 1;
      j++;
      x >>= 1;
    }
    if (!f)
      p[j]++;
    else
      b[j]++;
  }
  int l = 1, r = p[0], m, ans = -1;
  while (l <= r) {
    m = l + r >> 1;
    if (f(m)) {
      r = m - 1;
      ans = m;
    } else
      l = m + 1;
  }
  if (~ans) {
    for (int i = (ans); i <= (p[0]); i++) cout << i << " ";
    cout << endl;
  } else
    cout << ans << endl;
}
