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 n, i, j, k, s, a[50], b[50], c[50];
long long x;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%I64d", &x);
    for (j = 0; (1ll << j) < x; j++)
      ;
    if ((1ll << j) == x)
      a[j]++;
    else
      b[j - 1]++;
  }
  for (i = a[0]; i > 0; i--) {
    k = i;
    s = 0;
    for (j = 0; j <= 40; j++) {
      k = min(k, a[j]);
      c[j] = k;
    }
    for (j = 40; j >= 0; j--) {
      s += b[j] + a[j] - c[j];
      if (s > c[j]) break;
    }
    if (j >= 0) break;
  }
  if (i == a[0])
    cout << "-1";
  else
    for (i++; i <= a[0]; i++) cout << i << " ";
  return 0;
}
