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 double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const long long mod = (long long)(1e9 + 7);
const int MAX_N = 100010;
int main() {
  int n;
  while (~scanf("%d", &n)) {
    multiset<long long> ms1, ms2;
    for (int i = 1; i <= n; ++i) {
      long long t;
      scanf("%lld", &t);
      if (t == 1)
        ms1.insert(t);
      else if ((t & (t - 1)) == 0) {
        auto it = ms1.find(t >> 1);
        if (it != ms1.end()) {
          ms1.erase(it);
          ms1.insert(t);
        } else
          ms2.insert(t);
      } else
        ms2.insert(t);
    }
    int st = ms1.size();
    if (ms1.size() < ms2.size()) {
      puts("-1");
      continue;
    }
    if (ms2.size() > 0) {
      int flag = 0;
      auto tmp = ms2.end();
      for (--tmp;; --tmp) {
        long long now = *tmp;
        auto it2 = ms1.end();
        --it2;
        if (now <= ((*it2) << 1))
          ms1.erase(it2);
        else
          break;
        if (tmp == ms2.begin()) {
          flag = 1;
          break;
        }
      }
      if (!flag) {
        puts("-1");
        continue;
      }
    }
    vector<int> ans;
    ans.push_back(st);
    while (1) {
      st--;
      if (ms1.size() == 0) break;
      auto it = ms1.begin();
      long long val = *it;
      ms1.erase(it);
      int num = 0;
      for (int i = 0; i < 60; ++i) {
        if ((val >> i) & 1) {
          num = i + 1;
          break;
        }
      }
      if (ms1.size() < num) break;
      ans.push_back(st);
      for (int i = 0; i < num; ++i) {
        auto itt = ms1.end();
        --itt;
        ms1.erase(itt);
      }
    }
    reverse(ans.begin(), ans.end());
    for (int i : ans) {
      printf("%d ", i);
    }
    puts("");
  }
  return 0;
}
