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;
ifstream in("file.in");
ofstream out("file.out");
struct comp {
  bool operator()(long long a, long long b) { return a > b; }
};
long long n, first, t, r, a[100005], p[45], q[45], pos;
bool pw(long long first) {
  long long q = 1;
  while (!(first & q)) q *= 2;
  if (first - q) return 0;
  return 1;
}
int lg2(long long first) {
  long long q = 1, c = 0;
  while (first) {
    if (first & q) first -= q;
    q *= 2;
    c++;
  }
  return c - 1;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (pw(a[i]))
      p[lg2(a[i])]++;
    else
      q[lg2(a[i])]++;
  }
  for (int i = 1; i <= p[0]; i++) {
    t = i;
    r = 0;
    for (int j = 0; j <= 40; j++) {
      if (p[j] >= t)
        r += p[j] - t + q[j];
      else {
        r -= t - p[j];
        if (r < 0) r = 0;
        r += q[j];
        t = p[j];
      }
    }
    if (!r) cout << i << ' ', pos = 1;
  }
  if (!pos) cout << -1;
}
