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 MAX = 66;
int power[MAX + 10];
int between[MAX + 10];
int extra[MAX + 10];
int ende[MAX + 10];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long foo;
    cin >> foo;
    int cc = 0;
    bool good = true;
    while (foo > 1) {
      if (foo & 1) {
        good = false;
      }
      cc++;
      foo >>= 1;
    }
    if (good) {
      power[cc]++;
    } else {
      between[cc]++;
    }
  }
  int low = 1, high = power[0] + 1;
  while (low < high) {
    int mid = (low + high) >> 1;
    int cur = mid;
    for (int i = 0; i < MAX; i++) {
      extra[i] = power[i] - cur;
      int new_cur = min(cur, power[i + 1]);
      ende[i] = cur - new_cur;
      cur = new_cur;
    }
    bool fail = false;
    int open = 0;
    for (int i = MAX - 1; i >= 0; i--) {
      open += ende[i];
      open -= between[i];
      open -= extra[i + 1];
      if (open < 0) {
        fail = true;
        break;
      }
    }
    open -= extra[0];
    if (open < 0) {
      fail = true;
    }
    if (fail) {
      low = mid + 1;
    } else {
      high = mid;
    }
  }
  if (low > power[0]) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = low; i <= power[0]; i++) {
    cout << i << " ";
  }
  cout << endl;
  return 0;
}
