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 L = 45;
int n, cnt[L];
vector<long long> caps;
void impossible() {
  cout << -1 << '\n';
  exit(0);
}
int getPow(long long x) {
  int pow = 63 - __builtin_clzll(x);
  if (x != (1LL << pow)) {
    return -1;
  }
  return pow;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  long long a;
  for (int i = 0; i < n; i++) {
    cin >> a;
    int p2 = getPow(a);
    if (p2 == -1) {
      caps.emplace_back(a);
      continue;
    }
    cnt[p2]++;
  }
  for (int p = 1; p < L; p++) {
    while (cnt[p] > cnt[p - 1]) {
      caps.push_back(1LL << p);
      cnt[p]--;
    }
    cnt[p - 1] -= cnt[p];
  }
  sort(caps.rbegin(), caps.rend());
  vector<long long> towers;
  for (int p = 0; p < L; p++) {
    for (int i = 0; i < cnt[p]; i++) {
      towers.push_back(1LL << p);
    }
  }
  if (caps.size() > towers.size()) {
    impossible();
  }
  for (int i = 0; i < caps.size(); i++) {
    if (caps[i] > towers.back() * 2LL) {
      impossible();
    }
    towers.pop_back();
  }
  int mxsize = caps.size() + towers.size();
  int mnsize = mxsize;
  for (int i = 0; i < towers.size(); i++) {
    while (i != towers.size() && towers[i]) {
      towers.pop_back();
      towers[i] >>= 1;
    }
    if (i != towers.size() && !towers[i]) {
      mnsize -= 1;
    }
  }
  for (int i = mnsize; i <= mxsize; i++) {
    cout << i << ' ';
  }
  cout << '\n';
  return 0;
}
