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;
long long int index(long long int a) {
  long long int x = 1;
  long long int s = 0;
  while ((a & x) == 0) {
    ++s;
    x <<= 1;
  }
  return s;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<long long int> num(n);
  for (int i = 0; i < n; ++i) {
    cin >> num[i];
  }
  vector<long long int> ext;
  vector<int> count(60);
  vector<int> used(60);
  for (int i = 0; i < n; ++i) {
    if (((num[i] - 1) & num[i]) == 0) {
      ++count[index(num[i])];
    } else {
      ext.push_back(num[i]);
    }
  }
  int longest = 0;
  int M = 0;
  for (int i = 59; i >= 0; --i) {
    if (count[i] == 0) {
      continue;
    }
    M = max(M, i);
    int m = count[i];
    for (int j = i - 1; j >= 0; --j) {
      m = min(m, count[j]);
    }
    for (int j = 0; j < count[i] - m; ++j) {
      ext.push_back((1ll << i));
    }
    for (int j = i - 1; j >= 0; --j) {
      count[j] -= m;
    }
    count[i] = m;
    longest += m;
  }
  sort(ext.begin(), ext.end());
  int j = M;
  for (int i = ext.size() - 1; i >= 0; --i) {
    while ((j >= 0 && count[j] == 0)) {
      --j;
    }
    if ((1ll << (j + 1)) < ext[i]) {
      cout << -1 << endl;
      return 0;
    }
    --count[j];
    ++used[j];
  }
  long long int l = 0;
  long long int r = longest;
  while (l < r) {
    long long int m = l + (r - l) / 2;
    int acc = 0;
    int expend = 0;
    int i;
    int free = 0;
    for (i = 0; i <= M; ++i) {
      if (acc + count[i] <= m) {
        acc += count[i];
        expend += count[i] * (i + 1);
      } else {
        expend += (m - acc) * (i + 1);
        free += count[i] - (m - acc);
        break;
      }
      if (acc + used[i] <= m) {
        acc += used[i];
        expend += used[i] * (i + 2);
      } else {
        expend += (m - acc) * (i + 2);
        break;
      }
    }
    ++i;
    for (; i <= M; ++i) {
      free += count[i];
    }
    if (expend > free) {
      r = m;
    } else {
      l = m + 1;
    }
  }
  for (int i = longest - (l - 1); i <= longest; ++i) {
    cout << i << " ";
  }
  cout << endl;
}
