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 number[100005];
priority_queue<long long> q;
int power2[60];
int curr[60];
bool valid(int n, int m) {
  long long x;
  int ls = 0;
  while (q.size()) q.pop();
  if (!m) {
    return 0;
  }
  for (int i = 0; i < n; i++) {
    q.push(number[i]);
  }
  for (int i = 0; i < 44; i++) {
    if (power2[i] > m) {
      for (int j = 0; j < power2[i] - m; j++) {
        q.push((1LL << i));
      }
      curr[i] = m;
    } else {
      m = power2[i];
      curr[i] = power2[i];
    }
  }
  for (int i = 44; i >= 0; i--) {
    x = (1LL << i);
    curr[i] -= ls;
    ls += curr[i];
    while (curr[i] && q.size() && q.top() <= (x << 1)) {
      q.pop();
      curr[i]--;
    }
    if (curr[i] && q.size()) {
      return 0;
    }
  }
  return q.size() == 0;
}
int bs(int n) {
  int st = 0, ed = 100005;
  if (!valid(n, power2[0])) {
    return -1;
  }
  while (st < ed) {
    int mid = st + (ed - st + 1) / 2;
    if (!valid(n, mid)) {
      st = mid;
    } else {
      ed = mid - 1;
    }
  }
  return st + 1;
}
int main() {
  int n;
  long long a;
  int sz = 0;
  int answer = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a);
    for (int i = 0; i < 44; i++) {
      if ((1LL << i) == a) {
        power2[i]++;
        break;
      }
      if (i == 43) {
        number[sz++] = a;
      }
    }
  }
  answer = bs(sz);
  if (answer != -1) {
    for (int i = answer; i <= power2[0]; i++) {
      printf("%d ", i);
    }
  } else {
    printf("-1");
  }
  return 0;
}
