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;
int a[1000], b[1000], c[200000];
int main() {
  int n, x;
  long long t, p;
  cin >> n;
  x = 0;
  p = 1;
  a[-1] = 2147483647;
  for (int nn = 0; nn < n; ++nn) {
    cin >> t;
    if (t >= p * 2) {
      p *= 2;
      a[x - 1] -= a[x];
      x++;
    }
    if (t == p) {
      if (a[x] < a[x - 1])
        ++a[x];
      else
        ++b[x];
    } else
      ++b[x];
  }
  a[x - 1] -= a[x];
  int ans = 0;
  while (ans >= 0) {
    ans = 0;
    int tt = 0;
    for (int i = x; i >= 0; --i) {
      ans += a[i] - b[i];
      tt += a[i];
      if (ans < 0) break;
    }
    if (ans < 0) break;
    ++c[0];
    c[c[0]] = tt;
    int j = 0;
    while (a[j] == 0) ++j;
    --a[j];
    for (int i = 0; i <= j; ++i) ++b[i];
  }
  if (c[0] == 0)
    cout << "-1";
  else
    for (int i = c[0]; i > 0; --i) cout << c[i] << " ";
  cout << endl;
  return 0;
}
