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, num[101];
long long a;
int check(long long mid) {
  int mmin = mid, ret = num[0] - mid;
  for (int i = 1; i < 101; ++i) {
    ret = max(ret, 0);
    if (i & 1)
      ret += num[i];
    else {
      ret += (num[i] - mmin);
      mmin = min(mmin, num[i]);
    }
  }
  return !ret;
}
inline int id(long long x) {
  int ret = 0;
  while ((1LL << ret) < x) ++ret;
  if ((1LL << ret) == x) return ret << 1;
  return (ret << 1) - 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%I64d", &a);
    ++num[id(a)];
  }
  int left = 1, right = num[0], res = -1;
  while (left <= right) {
    int mid = (left + right) >> 1;
    if (check(mid)) {
      res = mid;
      right = mid - 1;
    } else
      left = mid + 1;
  }
  if (res == -1)
    puts("-1");
  else
    for (int i = res; i <= num[0]; ++i)
      printf("%d%c", i, i == num[0] ? '\n' : ' ');
  return 0;
}
