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;
int cnt[43];
set<long long> help;
map<long long, int> mp;
int cur[43];
int lft[43];
vector<int> fin;
int val(long long tmp) {
  int cn = 0;
  while (tmp) {
    tmp /= 2;
    cn++;
  }
  return cn;
}
int main() {
  int n;
  scanf("%d", &n);
  for (long long i = 1; i <= 2e12; i = i * 2) {
    help.insert(i);
    mp[i] = 0;
  }
  for (int i = 1; i <= n; i++) {
    long long a;
    scanf("%lld", &a);
    if (help.find(a) != help.end()) {
      mp[a]++;
    }
    cnt[val(a)]++;
  }
  int ans = 0;
  int num = 0;
  while (true) {
    long long lst = 0;
    for (map<long long, int>::iterator it = mp.begin(); it != mp.end(); it++) {
      if (it->second == 0) {
        break;
      } else {
        lst = it->first;
      }
    }
    if (lst == 0) {
      break;
    }
    num++;
    cur[val(lst)]++;
    int yu = 1;
    for (long long i = 1; i <= lst; i = i * 2) {
      cnt[yu]--;
      mp[i]--;
      yu++;
    }
    for (int i = 0; i <= 42; i++) {
      lft[i] = cur[i];
    }
    int flg = 0;
    for (int i = 42; i >= 1; i--) {
      if (i != 42) {
        lft[i] += lft[i + 1];
      }
      if (lft[i] >= cnt[i]) {
        lft[i] -= cnt[i];
      } else {
        flg = 1;
        break;
      }
    }
    if (flg == 0) {
      fin.push_back(num);
      ans++;
    }
  }
  if (ans == 0) {
    printf("-1\n");
  } else {
    for (int i = 0; i < fin.size(); i++) {
      printf("%d ", fin[i]);
    }
  }
  return 0;
}
