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>
#pragma GCC optimize("Ofast")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll getrnd(ll l, ll r) { return uniform_int_distribution<ll>(l, r)(rng); }
template <typename T1, typename T2>
inline bool relax(T1& a, const T2& b) {
  return a > b ? a = b, true : false;
}
template <typename T1, typename T2>
inline bool strain(T1& a, const T2& b) {
  return a < b ? a = b, true : false;
}
ll mxbit(ll x) {
  int len = 0;
  while (x) x /= 2, ++len;
  return len - 1;
}
void solve() {
  int n;
  cin >> n;
  const int LOG = 59;
  vector<ll> cnt(LOG), oth;
  for (int i = 0; i < n; ++i) {
    ll x;
    cin >> x;
    if (x == (1LL << mxbit(x)))
      ++cnt[mxbit(x)];
    else
      oth.emplace_back(x);
  }
  auto can = [&](int m) -> bool {
    ll mn = m, prev = 0;
    vector<ll> cp = cnt, en(LOG);
    for (int i = 0; i < LOG; ++i) {
      ll use = min(cp[i], mn);
      if (prev > use) en[i - 1] = prev - use;
      relax(mn, use);
      cp[i] -= use;
      prev = use;
    }
    for (ll i : oth) ++cp[mxbit(i) + 1];
    int lst = LOG - 1, tot = 0;
    for (int i = LOG - 1; i >= 0; --i) {
      tot += cp[i];
      while (cp[i]) {
        while (lst >= 0 && !en[lst]) --lst;
        if (lst == -1) return 0;
        if (lst + 1 < i) return 0;
        --cp[i];
        --en[lst];
      }
    }
    return tot <= m;
  };
  int l = -1, r = cnt[0] + 1;
  while (r - l > 1) {
    int mid = (l + r) / 2;
    if (can(mid))
      r = mid;
    else
      l = mid;
  }
  if (r > cnt[0]) cout << "-1";
  for (int i = r; i <= cnt[0]; ++i) cout << i << ' ';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  srand(time(0));
  int t = 1;
  while (t--) solve();
  return 0;
}
