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 n, i, f[100009], dem, d[50], sum, a[50];
vector<long long> t;
stack<long long> ans;
long long mu(int u, int v) {
  if (v == 0) return 1;
  return u * mu(u, v - 1);
}
long long kt(long long u) {
  long long tong = 0;
  while (u > 1) {
    if (u % 2 == 1) return 0;
    tong++;
    u /= 2;
  }
  return tong;
}
int main() {
  cin >> n;
  for (i = 0; i < n; i++) cin >> f[i];
  for (i = 0; i < n; i++) {
    if (f[i] == 1)
      d[0]++, a[0] = d[0];
    else {
      dem = kt(f[i]);
      if (dem == (long long)0)
        t.push_back(f[i]), sum++;
      else {
        if (d[dem] < d[dem - 1])
          d[dem]++, a[dem] = d[dem];
        else
          t.push_back(f[i]), sum++;
      }
    }
  }
  n = t.size();
  while (a[0] > 0) {
    for (i = 0;; i++) {
      if (a[i] == 0) break;
      a[i]--;
    }
    if (n > 0)
      if (t[n - 1] >= mu(2, i)) {
        cout << -1;
        return 0;
      }
    n--;
  }
  if (sum > d[0]) {
    cout << -1;
    return 0;
  }
  i = 0;
  dem = d[0];
  n = d[0];
  while (d[i] >= sum) {
    while (d[i] == d[i + 1]) {
      i++;
    }
    sum += (i + 1);
    d[i]--;
    dem--;
  }
  for (i = dem + 1; i <= n; i++) cout << i << ' ';
}
