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 nums[100005];
int nm[55], ns[55], n;
bool ok(int x) {
  int temp = 0;
  for (int i = 0; i < 50; i++) {
    if (x > ns[i]) {
      if (i)
        temp += nm[i] - nm[i - 1] - ns[i];
      else
        temp += nm[i] - ns[i];
      temp -= min(temp, x - ns[i]);
      x = ns[i];
    } else {
      if (i)
        temp += nm[i] - nm[i - 1] - x;
      else
        temp += nm[i] - x;
    }
  }
  return temp <= x;
}
int main() {
  bool flag = 0;
  cin >> n;
  int now = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", nums + i);
    while ((1LL << now) < nums[i]) {
      nm[now] = i - 1;
      now++;
    }
    if ((1LL << now) == nums[i]) ns[now]++;
  }
  while (now < 50) {
    nm[now] = n;
    now++;
  }
  for (int i = ns[0] / 2; i <= ns[0]; i++) {
    if (ok(i)) {
      if (flag)
        printf(" ");
      else
        flag = 1;
      printf("%d", i);
    }
  }
  if (!flag) printf("-1");
  printf("\n");
  return 0;
}
