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, a[100005], sum[50], cnt, an = -1, lol[100005], b[100005];
map<long long int, long long int> mp;
long long int binlo(long long int num) {
  long long int lo = 0, hi = n + 1, mid, tmp;
  while (lo <= hi) {
    mid = (lo + hi) >> 1;
    if (a[mid] < num)
      tmp = mid, lo = mid + 1;
    else
      hi = mid - 1;
  }
  return tmp;
}
long long int binup(long long int num) {
  long long int lo = 0, hi = n + 1, mid, tmp;
  while (lo <= hi) {
    mid = (lo + hi) >> 1;
    if (a[mid] > num)
      tmp = mid, hi = mid - 1;
    else
      lo = mid + 1;
  }
  return tmp;
}
bool check(long long int num) {
  long long int now = 1;
  memset(lol, false, sizeof(lol));
  lol[n + 1] = true, b[n + 1] = 0;
  for (long long int i = 1; i <= cnt; i++) {
    if (sum[i] > num)
      b[i] = num;
    else
      b[i] = sum[i];
  }
  long long int lef, ri, k;
  for (long long int i = 1; i <= cnt; i++) {
    lef = binlo(1LL << (i - 1)) + 1, ri = binup(1LL << (i - 1)), k = b[i];
    while (lef < ri && k) {
      if (!lol[lef]) k--, lol[lef] = true;
      lef++;
    }
    if (b[i] > b[i + 1]) {
      ri = binup(1LL << i), k = b[i];
      while (k > b[i + 1] && now < ri) {
        if (!lol[now]) k--, lol[now] = true;
        now++;
      }
    }
  }
  for (long long int i = 1; i <= n; i++) {
    if (!lol[i]) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  scanf("%lld", &n);
  for (long long int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    mp[a[i]]++;
  }
  sum[0] = 1000000007, a[0] = 0, a[n + 1] = (1LL << 45);
  for (long long int i = 1; i <= 40; i++) {
    sum[i] = min(sum[i - 1], mp[1LL << (i - 1)]);
    if (mp[1LL << (i - 1)] && sum[i] == 0) {
      printf("-1\n");
      return 0;
    }
    if (sum[i]) cnt = i;
  }
  long long int lo = 1, hi = sum[1], mid, tmp;
  while (lo <= hi) {
    mid = (lo + hi) >> 1;
    if (check(mid))
      an = mid, hi = mid - 1;
    else
      lo = mid + 1;
  }
  if (an == -1)
    printf("%lld\n", an);
  else
    for (long long int i = an; i <= sum[1]; i++) printf("%lld ", i);
  return 0;
}
