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 N = 55;
int n, m;
int rub[N], RUB[N], t[N], used[N];
bool sw;
bool ck(int x) {
  for (int i = 0; i < (m); i++) {
    used[i] = min(x, t[i]);
    x = used[i];
    RUB[i] = t[i] - used[i];
  }
  for (int i = (m)-1; i >= 0; i--)
    for (int j = 0; j < (i); j++) used[j] -= used[i];
  int ted = 0;
  for (int i = (m)-1; i >= 0; i--) {
    ted -= rub[i] + RUB[i + 1];
    if (ted < 0) return false;
    ted += used[i];
  }
  ted -= RUB[0];
  return (ted >= 0);
}
int f(long long x) {
  int cnt = 0;
  while (x) {
    cnt++;
    x /= 2;
  }
  m = max(m, cnt + 1);
  cnt--;
  return cnt;
}
int main() {
  cin >> n;
  for (int i = 0; i < (n); i++) {
    long long x, y;
    cin >> x;
    y = f(x);
    if ((1LL << y) == x)
      t[y]++;
    else
      rub[y - 1]++;
  }
  for (int i = 0; i <= t[0]; i++)
    if (ck(i)) cout << i << " ", sw = true;
  if (!sw) cout << -1;
  return 0;
}
