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;
vector<long long> vec;
int n;
vector<int> a, b, res;
bool possible(int t) {
  int tinit = t;
  vector<int> aux = a, bux = b;
  if (aux[0] < t) {
    return false;
  }
  bux[1] += aux[0] - t;
  for (int i = 1; i < 41; i++) {
    if (bux[i] > t) {
      return false;
    }
    if (aux[i] > t) {
      bux[i + 1] += aux[i] - t;
    } else if (aux[i] < t) {
      bux[i] -= t - aux[i];
      bux[i] = max(bux[i], 0);
      t = aux[i];
    }
    bux[i + 1] += bux[i];
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  vec.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> vec[i];
  }
  a.resize(42);
  b.resize(42);
  for (int i = 0; i < 42; i++) {
    a[i] = 0;
    b[i] = 0;
  }
  int pot = 0;
  for (int i = 0; i < vec.size(); i++) {
    while (vec[i] > (1LL << pot)) {
      pot++;
    }
    if (vec[i] == (1LL << pot)) {
      a[pot]++;
    } else {
      b[pot]++;
    }
  }
  res.clear();
  for (int i = 1; i <= n; i++) {
    if (possible(i)) {
      res.push_back(i);
    }
  }
  if (res.empty()) {
    cout << -1 << endl;
  } else {
    for (int i = 0; i < res.size(); i++) {
      if (i == res.size() - 1) {
        cout << res[i] << endl;
      } else {
        cout << res[i] << " ";
      }
    }
  }
  return 0;
}
