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;
long long int N = 40;
long long int power[41] = {0};
vector<long long int> dpg(41, 0);
vector<long long int> cnt(41, 0);
map<long long int, long long int> mp;
bool ok(long long int k) {
  if (cnt[0] < k) return false;
  vector<long long int> dp = dpg;
  long long int rem = dp[0] + cnt[0];
  long long int last = k;
  vector<long long int> l(N + 1);
  for (long long int j = 0; j <= N; j++) {
    last = min(cnt[j], last);
    for (long long int i = 0; i < j; i++) {
      dp[i] -= last;
    }
    rem -= last;
    l[j] = last;
  }
  for (long long int j = N; j >= 0; j--) {
    if (dp[j] > l[j]) {
      return false;
    }
  }
  if (rem > k) {
    return false;
  }
  return true;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  for (long long int j = 0; j < n; j++) {
    cin >> v[j];
    if (mp.find(v[j]) != mp.end()) {
      cnt[mp[v[j]]]++;
    }
    long long int i = 0;
    while (power[i] < v[j]) {
      dpg[i]++;
      i++;
    }
  }
  bool lol = false;
  for (long long int k = 1; k <= n; k++) {
    if (ok(k)) {
      lol = true;
      cout << k << " ";
    }
  }
  if (!lol) {
    cout << -1;
  }
  cout << "\n";
  return;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  t = 1;
  power[0] = 1;
  mp[1] = 0;
  for (long long int j = 1; j <= 40; j++) {
    power[j] = power[j - 1] * 2;
    mp[power[j]] = j;
  }
  while (t--) {
    solve();
  }
  return 0;
}
