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;
  vector<long long> C(1), other;
  int oCnt = 0;
  long long temp;
  long long curr = 1;
  for (int i = 0; i < N; i++) {
    cin >> temp;
    while (temp > curr) {
      curr *= 2;
      C.push_back(0);
    }
    if (temp == curr)
      C.back()++;
    else if (temp < curr) {
      oCnt++;
      other.push_back(temp);
    }
  }
  for (int i = 1; i < C.size(); i++) {
    if (C[i] > C[i - 1]) {
      oCnt += C[i] - C[i - 1];
      for (int j = 0; j < C[i] - C[i - 1]; j++)
        other.push_back((long long)1 << i);
      C[i] = C[i - 1];
    }
  }
  while (!C.empty() and C.back() == 0) C.pop_back();
  if (C.empty() or C[0] < oCnt) {
    cout << -1 << endl;
    return 0;
  }
  sort(other.begin(), other.end());
  for (int i = C.size() - 1; i >= 0; i--) {
    int extra = (i + 1 == C.size()) ? C[i] : C[i] - C[i + 1];
    for (int j = 0; j < extra; j++) {
      long long temp = ((long long)1 << (i + 1));
      if (!other.empty() and temp >= other.back()) other.pop_back();
    }
  }
  if (!other.empty()) {
    cout << -1 << endl;
    return 0;
  }
  int maxM = C[0];
  int minM = 0;
  while (!C.empty()) {
    minM++;
    for (int i = 0; i < C.size(); i++) C[i]--;
    while (!C.empty() and C.back() == 0) C.pop_back();
    if (oCnt > 0)
      oCnt--;
    else {
      for (int i = 0; i < C.size(); i++) {
        if (i + 1 == C.size() or C[i] > C[i + 1]) {
          C[i]--;
          break;
        }
      }
    }
    while (!C.empty() and C.back() == 0) C.pop_back();
  }
  for (int i = minM; i <= maxM; i++) cout << i << (i != maxM ? " " : "");
  cout << endl;
}
