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 ll = long long;
using ld = long double;
using ull = unsigned long long;
using namespace std;
const int N = 1e5 + 2;
int n, cnt[40], cnt1 = 0, mx;
ll a[N];
vector<ll> b;
bool ok(int num) {
  int cur = num;
  vector<ll> redun = b;
  vector<pair<int, ll> > lst;
  for (int i = (0); i <= (mx); ++i) {
    if (cur > cnt[i]) {
      lst.emplace_back(cur - cnt[i], (1LL << (i - 1)));
      cur = cnt[i];
    } else
      for (int j = (1); j <= (cnt[i] - cur); ++j) redun.emplace_back(1LL << i);
  }
  if (cur) lst.emplace_back(cur, (1LL << mx));
  if (((int)(redun).size()) > num) return 0;
  sort((redun).begin(), (redun).end(), greater<ll>());
  if (redun.empty()) return 1;
  int i = 0;
  reverse((lst).begin(), (lst).end());
  for (pair<int, ll> v : lst) {
    if (redun[i] >= v.second * 2) return 0;
    i += v.first;
    if (i >= ((int)(redun).size())) break;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = (1); i <= (n); ++i) {
    cin >> a[i];
    int x = __builtin_ctzll(a[i]);
    if (a[i] == (1LL << x))
      cnt[x]++, mx = max(mx, x);
    else
      b.emplace_back(a[i]);
  }
  int num = 0;
  cnt1 = cnt[0];
  int l = ((int)(b).size()), r = cnt1;
  while (l <= r) {
    int m = (l + r) >> 1;
    if (ok(m))
      r = m - 1;
    else
      l = m + 1;
  }
  if (l > cnt1)
    cout << -1;
  else
    for (int i = (l); i <= (cnt1); ++i) cout << i << ' ';
  return 0;
}
