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 N;
long long A[100005];
vector<long long> X;
int C[45];
long long P[45];
void power() {
  P[0] = 1;
  for (int i = 1; i <= 40; i++) P[i] = P[i - 1] * 2;
}
int chk(int S) {
  vector<long long> tmp;
  vector<long long> T;
  int L = X.size();
  for (int i = 0; i < L; i++) tmp.push_back(X[i]);
  int c[45];
  for (int i = 0; i < 45; i++) c[i] = C[i];
  int curr = 0;
  for (int i = 40; i >= 0; i--) {
    int minimum = INT_MAX;
    for (int j = 0; j <= i; j++) {
      if (c[j] < minimum) minimum = c[j];
    }
    minimum = min(minimum, (S - curr));
    for (int j = 0; j <= i; j++) c[j] -= minimum;
    for (int j = 1; j <= minimum; j++) T.push_back(P[i]);
    curr += minimum;
  }
  for (int i = 0; i <= 40; i++) {
    for (int j = 1; j <= c[i]; j++) tmp.push_back(P[i]);
  }
  sort(tmp.begin(), tmp.end());
  sort(T.begin(), T.end());
  int L1 = T.size();
  int L2 = tmp.size();
  if (L2 > L1) return 0;
  int ccc = L1 - 1;
  for (int i = L2 - 1; i >= 0; i--) {
    if ((2 * T[ccc]) < tmp[i]) return 0;
    ccc--;
  }
  return 1;
}
void find(long long S) {
  long long prod = 1;
  int cnt = 0;
  while (prod <= S) {
    if (prod == S) {
      C[cnt]++;
      return;
    }
    prod *= 2;
    cnt++;
  }
  X.push_back(S);
  return;
}
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> A[i];
  for (int i = 0; i < N; i++) find(A[i]);
  power();
  int low = 0;
  int high = C[0];
  int mid;
  int pos = -1;
  while (high >= low) {
    mid = (high + low) / 2;
    if (chk(mid)) {
      pos = mid;
      high = mid - 1;
    } else
      low = mid + 1;
  }
  if (pos == -1)
    cout << "-1";
  else
    for (int i = pos; i <= C[0]; i++) cout << i << " ";
  cout << endl;
  return 0;
}
