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;
int a[239], b[239];
vector<int> R;
bool judge(int x) {
  memcpy(b, a, sizeof(b));
  map<int, int> del;
  for (int j = (1); j < x + 1; j++) {
    if (!b[0]) break;
    for (int i = (0); i < 60; i++) {
      b[i]--;
      if (!b[i + 1]) {
        del[i]++;
        break;
      }
    }
  }
  vector<int> nd(R);
  for (int i = (0); i < 239; i++)
    while (b[i]) {
      nd.push_back(i);
      b[i]--;
    }
  for (auto p : nd) {
    auto iter = del.lower_bound(p);
    if (iter == del.end() || !(iter->second)) return false;
    iter->second--;
    if (!(iter->second)) del.erase(iter);
  }
  return true;
}
int Bsearch(int l, int r) {
  while (l < r) {
    long long mid = (l + r) >> 1;
    if (judge(mid))
      r = mid;
    else
      l = mid + 1;
  }
  return l;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i < n + 1; i++) {
    long long x;
    scanf("%lld", &x);
    if (((x) & (-x)) == x) {
      int lll = 0;
      while (x / 2) {
        x >>= 1;
        lll++;
      }
      a[lll]++;
    } else
      R.push_back(log2(x));
  }
  if (!judge(a[0]))
    printf("-1\n");
  else {
    int x = Bsearch(1, a[0]);
    for (int i = (x); i < a[0] + 1; i++) printf("%d ", i);
  }
  return 0;
}
