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;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
map<long long, int> T;
vector<long long> R, P, o;
int N, J;
long long A[(1 << 17)];
long long gt() {
  long long p = 1;
  while (T[p]) --T[p], p <<= 1;
  return p / 2;
}
int main(void) {
  scanf("%d", &N);
  for (int i(0); i < N; i++) scanf("%lld", A + i);
  for (int i(0); i < N; i++)
    if (!(!(A[i] & (A[i] - 1))))
      R.push_back(A[i]);
    else
      ++T[A[i]];
  while (T[1]) P.push_back(gt());
  for (auto &h : T)
    for (int i(0); i < h.second; i++) R.push_back(h.first);
  sort(R.begin(), R.end(), greater<long long>()), J = P.size(),
                                                  sort(P.begin(), P.end());
  if (R.size() > P.size()) return puts("-1");
  for (auto &h : R)
    if (h >= P.back() * 2)
      return puts("-1");
    else
      P.pop_back();
  o.push_back(J);
  for (int i(0); i < (int)P.size(); i++) {
    while (i < (int)P.size() && P[i]) {
      if (P[i] > P.back() * 2) goto GOW;
      P[i] /= 2, P.pop_back();
    }
    if (i < (int)P.size()) o.push_back(--J);
  }
GOW:
  sort(o.begin(), o.end());
  for (int i(0); i < (int)o.size(); i++)
    printf("%lld%c", o[i], i + 1 == (int)o.size() ? 10 : 32);
  return 0;
}
