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;
int bases[40 + 1], tails[40 + 1];
int minLen, maxLen;
int log2(long long n) {
  if (n <= 1) return (int)n;
  return 1 + log2(n >> 1);
}
bool calc() {
  for (int i = 2; i <= 40; i++) {
    if (bases[i] > bases[i - 1]) {
      tails[i - 1] += bases[i] - bases[i - 1];
      bases[i] = bases[i - 1];
    }
  }
  for (int i = 40; i > 0; i--) {
    if (!tails[i]) continue;
    if (bases[i] < tails[i]) return false;
    for (int j = 1; j <= i; j++) {
      bases[j] -= tails[i];
    }
    for (int j = 0; j < tails[i]; j++) {
      for (int k = i + 1; k <= 40; k++) {
        if (bases[k] > 0)
          bases[k]--;
        else
          break;
      }
    }
    minLen += tails[i];
  }
  maxLen = minLen + bases[1];
  int tailFree = bases[1];
  int currBase = 1;
  while (tailFree > 0) {
    for (int i = currBase + 1; i <= 40; i++) {
      if (bases[i] == bases[currBase])
        currBase++;
      else
        break;
    }
    if (tailFree - 1 >= currBase) {
      tailFree -= currBase + 1;
      bases[currBase]--;
    } else
      break;
  }
  minLen += bases[currBase];
  return true;
}
int main() {
  int n;
  scanf("%d\n", &n);
  for (int i = 0; i < n; i++) {
    long long ai;
    scanf("%lld", &ai);
    int order = log2(ai);
    if (ai == 1LL << (order - 1))
      bases[order]++;
    else
      tails[order]++;
  }
  if (!calc()) {
    printf("-1\n");
  } else {
    for (int i = minLen; i <= maxLen; i++) printf("%d ", i);
    printf("\n");
  }
  return 0;
}
