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 long long int MOD = 1000000007;
long long int N;
long long int num[100000];
long long int on[50];
long long int over[50];
long long int box[50];
list<long long int> ans;
int main() {
  ios::sync_with_stdio(false);
  box[0] = 1;
  for (int i = 1; i < 50; i++) {
    box[i] = box[i - 1] * 2;
  }
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> num[i];
    for (int j = 49; j >= 0; j--) {
      if (num[i] == box[j]) {
        on[j]++;
        break;
      }
      if (num[i] > box[j]) {
        over[j]++;
        break;
      }
    }
  }
  for (int i = 0; i < 49; i++) {
    if (on[i] < on[i + 1]) {
      over[i + 1] += on[i + 1] - on[i];
      on[i + 1] = on[i];
    }
  }
  for (long long int i = (on[0] + 1) / 2; i <= on[0]; i++) {
    long long int box_on[50] = {};
    long long int box_over[50] = {};
    long long int s_over[50] = {};
    for (int j = 0; j < 50; j++) {
      box_over[j] = over[j];
      if (on[j] > i) {
        box_on[j] = i;
        box_over[j] += on[j] - i;
      } else {
        box_on[j] = on[j];
      }
    }
    s_over[49] = box_over[49];
    bool flag = true;
    for (int j = 48; j >= 0; j--) {
      s_over[j] = s_over[j + 1] + box_over[j];
      if (s_over[j] > box_on[j]) {
        flag = false;
        break;
      }
    }
    if (flag) ans.push_back(i);
  }
  if (ans.empty()) {
    cout << "-1\n";
    return 0;
  }
  for (auto i : ans) {
    cout << i << " ";
  }
  cout << endl;
  return 0;
}
