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 = 0x3f3f3f3f;
const double eps = 1e-8;
long long int a[100005];
vector<long long int> v;
vector<int> ans;
map<long long int, int> s;
long long int two[60];
int num[60], used[60];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  two[0] = 1;
  for (int i = 0; i < 60; ++i) {
    s[two[i]] = i;
    two[i + 1] = two[i] << 1;
  }
  while (cin >> n) {
    memset(num, 0, sizeof(num));
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      if (s.count(a[i])) {
        num[s[a[i]]]++;
      } else {
        v.push_back(a[i]);
      }
    }
    int mx = 0, fre, idx, tmp, mn = 0;
    for (int i = 0; i < 60; ++i)
      if (num[i]) mx = i;
    bool good = num[0] >= v.size();
    for (int i = 1; i < 60 && good; ++i) {
      if (num[i] > 2 * num[i - 1]) good = 0;
    }
    bool bad = 0;
    for (int i = (num[0] + 1) / 2; i <= num[0] && good; ++i) {
      memcpy(used, num, (mx + 1) * sizeof(int));
      fre = i, idx = mn = 0, bad = 0;
      used[0] -= i;
      if (!used[0]) mn = 1;
      for (int j = 1; j <= mx && !bad; ++j) {
        if (used[j] < fre) {
          while (idx < v.size() && fre > used[j]) {
            if (v[idx] < two[j]) {
              --fre;
              ++idx;
            } else
              break;
          }
          if (fre > used[j]) {
            for (int k = mn; k < j; ++k) {
              tmp = fre - used[j];
              if (tmp >= used[k]) {
                fre -= used[k];
                used[k] = 0;
                ++mn;
              } else {
                used[k] -= tmp;
                fre = used[j];
              }
              if (fre == used[j]) break;
            }
            fre = used[j];
          }
        }
        used[j] -= fre;
        if (mn == j && !used[j]) ++mn;
      }
      while (idx < v.size() && fre > 0) {
        if (v[idx] < two[mx + 1]) {
          ++idx;
          --fre;
        } else
          break;
      }
      for (int i = mn; i <= mx; ++i) {
        if (used[i]) {
          if (fre >= used[i]) {
            fre -= used[i];
            used[i] = 0;
          } else
            bad = 1;
        }
      }
      if (!bad && idx == v.size()) ans.push_back(i);
    }
    if (!good || !ans.size())
      cout << "-1\n";
    else {
      for (int i = 0; i < ans.size(); ++i) cout << ans[i] << " ";
      cout << "\n";
    }
    v.clear();
    ans.clear();
  }
  return 0;
}
