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 maxn = 1e5 + 10;
long long a[maxn];
int n, cnt[40];
vector<long long> rem;
bool v[maxn];
bool check(int);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
  for (int p = 0; p < 40; ++p) {
    auto it = lower_bound(a, a + n, (1LL << p));
    if (*it != (1LL << p)) continue;
    auto itt = upper_bound(a, a + n, (1LL << p));
    cnt[p] = itt - it;
    for (int i = it - a; i < itt - a; ++i) v[i] = true;
  }
  for (int i = 0; i < n; ++i)
    if (!v[i]) rem.push_back(a[i]);
  int d = 1, ans = cnt[0] + 1;
  while (d <= cnt[0]) d <<= 1;
  while (d >>= 1)
    if (ans - d >= 0)
      if (check(ans - d)) ans -= d;
  if (!check(ans)) return cout << "-1\n", 0;
  for (int i = ans;; ++i) {
    if (cnt[0] >= i)
      cout << i << ' ';
    else
      break;
  }
  return cout << '\n', 0;
}
bool check(int m) {
  vector<int> len, c;
  for (int i = 0; i < 40; ++i) c.push_back(cnt[i]);
  for (int i = 0; i < m; ++i) {
    if (!c[0]) return false;
    --c[0];
    int cur = 0;
    for (int j = 1; j < 40; ++j) {
      if (c[j])
        ++cur, --c[j];
      else
        break;
    }
    len.push_back(cur);
  }
  if (len.size() < m) return false;
  priority_queue<long long, vector<long long>, less<long long>> pq(rem.begin(),
                                                                   rem.end());
  for (int i = 0; i < 40; ++i) {
    while (c[i]--) pq.push((1LL << i));
  }
  if (pq.size() > len.size()) return false;
  int ind = 0;
  while (pq.size() && ind < len.size()) {
    if (pq.top() > (1LL << (len[ind] + 1))) return false;
    pq.pop();
    ++ind;
  }
  return true;
}
