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 o[100010];
long long pw[60];
priority_queue<long long> pq;
vector<long long> k[100010];
int main() {
  long long i, j, n;
  long long cnt;
  long long sz;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &o[i]);
    if ((o[i] & (-o[i])) != o[i])
      pq.push(o[i]);
    else {
      long long tmp = o[i];
      cnt = 0;
      while (tmp > 1) {
        tmp >>= 1;
        cnt++;
      }
      pw[cnt]++;
    }
  }
  sz = pw[0];
  for (i = 0; i < sz; i++) k[i].push_back(1);
  for (i = 1; i < 60; i++) {
    if (pw[i] == 0) continue;
    cnt = 0;
    while (pw[i] > 0 && cnt < sz) {
      if (k[cnt].back() * 2 == ((long long)1 << i))
        k[cnt].push_back((long long)1 << i);
      else
        break;
      cnt++;
      pw[i]--;
    }
    while (pw[i] > 0) {
      pq.push((long long)1 << i);
      pw[i]--;
    }
  }
  long long ss = pq.size();
  if (pq.size() > 0) {
    for (i = 0; i < sz; i++) {
      long long tmp = pq.top();
      pq.pop();
      if (tmp <= k[i].back() * 2)
        k[i].push_back(tmp);
      else {
        printf("-1\n");
        return 0;
      }
      if (pq.size() == 0) break;
    }
    if (pq.size()) {
      printf("-1\n");
      return 0;
    }
  }
  vector<long long> ans;
  long long all = sz - ss;
  ans.push_back(sz);
  for (i = sz - 1; i >= 0; i--) {
    if (all >= k[i].size() + 1) {
      for (auto v : k[i]) pq.push(v);
      long long ff = pq.size();
      for (j = ss; j < sz; j++) {
        long long tmp = pq.top();
        if (tmp <= k[j].back() * 2)
          k[j].push_back(tmp);
        else
          break;
        pq.pop();
        if (pq.size() == 0) break;
      }
      if (pq.size() > 0)
        break;
      else
        ss += ff, all -= ff + 1;
      ans.push_back(i);
    } else
      break;
  }
  for (i = ans.size() - 1; i >= 0; i--) printf("%lld ", ans[i]);
  printf("\n");
  return 0;
}
