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;
long long a[100010];
bool check(int mid) {
  int sum = 0, j = 1;
  for (long long i = 0; i <= 40; i++) {
    long long now = (1LL << i);
    int cnt = 0;
    for (; j <= n && a[j] <= now; j++) {
      if (a[j] == now)
        cnt++;
      else if (a[j] < now)
        sum++;
    }
    if (sum > mid) return false;
    if (mid > cnt) {
      sum -= (mid - cnt);
      if (sum < 0) sum = 0;
      mid = cnt;
    } else if (cnt >= mid)
      sum += cnt - mid;
    if (sum > mid) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  int R = 0;
  for (int i = 1; i < (n + 1); i++) scanf("%I64d", &a[i]);
  for (int i = 1; i < (n + 1); i++) R += (a[i] == 1 ? 1 : 0);
  int tmpR = R;
  int L = 1, ans = R + 1;
  while (L <= R) {
    int mid = ((L + R) >> 1);
    if (check(mid)) {
      R = mid - 1;
      ans = mid;
    } else
      L = mid + 1;
  }
  if (ans > tmpR)
    puts("-1");
  else {
    for (int i = ans; i < (tmpR + 1); i++) printf("%d ", i);
    puts("");
  }
}
