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;
map<long long, int> M;
vector<vector<long long> > v;
int main() {
  int n;
  cin >> n;
  multiset<long long> m;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    m.insert(x);
  }
  while (1) {
    if (m.find(1) == m.end()) {
      break;
    }
    vector<long long> v2;
    for (long long i = 1;; i *= 2) {
      if (m.find(i) != m.end()) {
        v2.push_back(i);
        m.erase(m.find(i));
      } else {
        break;
      }
    }
    v.push_back(v2);
  }
  int lo = 0, hi = (int)v.size() + 1, mi;
  while (hi - lo > 1) {
    mi = (lo + hi) >> 1;
    bool ok = 1;
    int cnt = (int)m.size();
    multiset<long long> m2 = m;
    for (int i = mi; i < (int)v.size(); i++) {
      cnt += (int)v[i].size();
      for (int j = 0; j < (int)v[i].size(); j++) {
        m2.insert(v[i][j]);
      }
    }
    if (cnt > mi) ok = 0;
    int pos = 0;
    if ((int)m2.size() && ok) {
      multiset<long long>::iterator it = m2.end();
      it--;
      for (;; it--) {
        if (v[pos][(int)v[pos].size() - 1] * 2 < *it) {
          ok = 0;
        }
        if (it == m2.begin()) break;
        pos++;
      }
    }
    if (ok) {
      hi = mi;
    } else
      lo = mi;
  }
  if (hi == (int)v.size() + 1) {
    cout << -1 << endl;
  } else {
    for (int i = 0; i < (int)v.size() - hi + 1; i++) {
      if (i) cout << " ";
      cout << (hi + i);
    }
    cout << endl;
  }
}
