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;
const long long inf = (long long)1e18;
const long long mod = (long long)1e9 + 7;
const double pi = acos(-1.0);
const double eps = (double)1e-9;
const int dx[] = {0, 0, 1, 0, -1};
const int dy[] = {0, 1, 0, -1, 0};
const int N = 100500;
vector<int> ans;
int n, a[45], b[45];
inline bool check(int x) {
  int c[45], d[45], end[45];
  memset(end, 0, sizeof(end));
  for (int i = 0; i <= 40; i++) {
    c[i] = a[i];
    d[i] = b[i];
  }
  int mn = min(c[0], x);
  if (mn < x) return 0;
  int pr = mn;
  for (int i = 0; i <= 40; i++) {
    mn = min(mn, c[i]);
    end[i] = pr - mn;
    c[i] -= mn;
    pr = mn;
  }
  int cur = 0;
  for (int i = 0; i <= 40; i++) {
    cur += c[i] + d[i];
    mn = min(cur, end[i]);
    cur -= mn;
    end[i] -= mn;
  }
  return cur == 0;
}
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    long long y = 1;
    int pw = 0;
    while (y < x) y <<= 1, pw++;
    if (x == y)
      a[pw]++;
    else
      b[pw]++;
  }
  for (int i = 1; i <= n; i++)
    if (check(i)) ans.push_back(i);
  if (ans.empty()) ans.push_back(-1);
  for (auto i : ans) cout << i << " ";
  return 0;
}
