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 inf = 2 * 1e9;
const int maxn = 45;
int n;
int stcnt[maxn], staft[maxn], nwstcnt[maxn], nwtoclose[maxn], enders[maxn];
bool check(int mx) {
  int nowforend = 0;
  for (int i = 0; i <= 42; ++i) {
    int nextmx = mx;
    if (stcnt[i + 1] < nextmx) nextmx = stcnt[i + 1];
    enders[i] = stcnt[i] - mx;
    nwtoclose[i] = mx - nextmx;
    mx = nextmx;
  }
  for (int i = 42; i >= 0; --i) {
    nowforend = nowforend + nwtoclose[i] - staft[i] - enders[i];
    if (nowforend < 0) return false;
  }
  if (nowforend < 0) return false;
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  long long st = 1;
  int step = 0;
  for (int i = 1; i <= n; ++i) {
    long long a;
    cin >> a;
    while (a >= st * 2) st *= 2, step += 1;
    if (a == st)
      stcnt[step] += 1;
    else
      staft[step] += 1;
  }
  int l = 0, r = stcnt[0];
  while (r - l > 1) {
    int md = l + (r - l) / 2;
    if (check(md))
      r = md;
    else
      l = md;
  }
  if (!check(r))
    cout << -1;
  else {
    for (int i = r; i <= stcnt[0]; ++i) cout << i << " ";
  }
  return 0;
}
