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 num[45], r[45], ans[100010], p, n, tot;
long long x, tmp;
bool check(int x) {
  int rest = num[0] - x;
  for (int i = 1; i <= p; i++) {
    if (x > num[i]) {
      rest -= min(x - num[i], rest);
      x = num[i];
    } else
      rest += num[i] - x;
    rest += r[i];
  }
  return rest <= x;
}
int main() {
  scanf("%d", &n);
  tmp = 1;
  p = 0;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (x == tmp)
      num[p]++;
    else if (x < tmp)
      r[p - 1]++;
    else {
      for (; tmp < x;) {
        tmp <<= 1;
        p++;
      }
      if (x == tmp)
        num[p]++;
      else
        r[p - 1]++;
    }
  }
  for (int i = 1; i <= num[0]; i++)
    if (check(i)) ans[++tot] = i;
  if (tot) {
    for (int i = 1; i < tot; i++) printf("%d ", ans[i]);
    printf("%d", ans[tot]);
  } else
    printf("-1");
  return 0;
}
