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;
const int K = 50;
int a[K], b[K];
int tz(long long t) { return t == 1 ? 0 : 1 + tz(t >> 1); }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    long long t;
    scanf("%lld", &t);
    if (t == (t & -t))
      a[tz(t)]++;
    else
      b[tz(t)]++;
  }
  vector<int> ans;
  for (;;) {
    for (int i = 1; i < K; i++)
      if (a[i - 1] < a[i]) {
        b[i] += a[i] - a[i - 1];
        a[i] = a[i - 1];
      }
    int cnt = 0;
    for (int i = K - 2; i >= 0; i--) {
      cnt += a[i] - a[i + 1];
      if (cnt < b[i]) {
        cnt = -1;
        break;
      }
      cnt -= b[i];
    }
    if (cnt == -1) break;
    ans.push_back(a[0]);
    a[0]--, b[0]++;
  }
  reverse(ans.begin(), ans.end());
  if (ans.size() == 0) ans.push_back(-1);
  for (int x : ans) printf("%d ", x);
  printf("\n");
}
