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;
int main() {
  int n;
  cin >> n;
  int no[41];
  int rn[41];
  for (int i = 0; i <= 40; i++) {
    no[i] = 0;
    rn[i] = 0;
  }
  long long cv = 1;
  int noi = 0;
  int ctr = 0;
  long long nm;
  while (ctr < n) {
    cin >> nm;
    ctr++;
    while (nm >= (cv * 2)) {
      cv *= 2;
      noi++;
    }
    if (nm == cv && nm == 1) {
      no[noi]++;
      continue;
    }
    if (nm == cv && no[noi] < no[noi - 1]) {
      no[noi]++;
      continue;
    }
    rn[noi]++;
  }
  int nc = 0;
  int nen = 0;
  for (int i = 40; i >= 0; i--) {
    nc = no[i];
    nen += rn[i];
    if (nc < nen) {
      cout << -1 << endl;
      return 0;
    }
  }
  int mnn = no[0];
  int cn = 0;
  while (no[0] > 0) {
    cn++;
    int ic = 0;
    for (; ic <= 40; ic++) {
      if (no[ic] > 0) {
        no[ic]--;
        continue;
      }
      break;
    }
    ic--;
    bool jmu = 0;
    for (int i = ic; i >= 0; i--) {
      if (rn[i] > 0) {
        rn[i]--;
        jmu = 1;
        break;
      }
    }
    if (jmu) {
      continue;
    }
    if (!jmu) {
      if (no[0] == 0) {
        break;
      }
      for (ic = 0; ic <= 40; ic++) {
        no[ic]--;
        rn[ic]++;
        if (ic < 40 && no[ic] < no[ic + 1]) {
          continue;
        }
        break;
      }
    }
    rn[ic]--;
  }
  if (rn[0] > 0) {
    cn++;
  }
  for (int i = cn; i <= mnn; i++) {
    if (i > cn) {
      cout << " ";
    }
    cout << i;
  }
  cout << endl;
  cin >> n;
}
