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;
long long n, x, a[100005], dp[100005], b[100005], f[45], l, r, mid,
    ans(1e18), now;
map<long long, int> m;
vector<long long> v, w;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  f[0] = 1;
  m[1] = 0;
  for (int i = 1; i <= 41; i++) {
    f[i] = f[i - 1] * 2;
    m[f[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (m.count(x)) {
      a[m[x]]++;
    } else {
      v.push_back(x);
    }
  }
  l = max(1ll, a[0] / 2), r = a[0];
  while (l <= r) {
    mid = (l + r) >> 1;
    w = v, now = mid;
    for (int i = 0; i <= 40; i++) {
      dp[i] = a[i];
    }
    for (int i = 1; i <= mid; i++) {
      b[i] = -1;
    }
    for (int i = 0; i <= 40; i++) {
      for (int j = 1; j <= now; j++) {
        if (!dp[i]) {
          now = j - 1;
          break;
        } else {
          dp[i]--;
          b[j] = i;
        }
      }
    }
    for (int i = 0; i <= 40; i++) {
      for (int j = 0; j < dp[i]; j++) {
        w.push_back(f[i]);
      }
    }
    sort(w.begin(), w.end());
    now = 1;
    while (w.size()) {
      if (w.back() <= f[b[now] + 1]) {
        now++;
      } else {
        break;
      }
      w.pop_back();
      if (now > mid) {
        break;
      }
    }
    if (w.size()) {
      l = mid + 1;
    } else {
      r = mid - 1;
      ans = mid;
    }
  }
  if (ans == 1e18) {
    cout << -1 << '\n';
    return 0;
  }
  for (int i = ans; i <= a[0]; i++) {
    cout << i << " ";
  }
  cout << '\n';
}
