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;
const int N = 77;
const int MAX = 55;
long long full[N];
long long notfull[N];
long long extra[N];
long long ende[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    long long foo;
    scanf("%I64d", &foo);
    bool bar = true;
    int use = 0;
    while (foo > 1) {
      if (foo & 1) {
        bar = false;
      }
      foo >>= 1;
      use++;
    }
    if (bar) {
      full[use]++;
    } else {
      notfull[use]++;
    }
  }
  int low = 1, high = full[0] + 1;
  while (low < high) {
    int mid = (low + high) >> 1;
    int cur = mid;
    for (int i = 0; i < MAX; i++) {
      extra[i] = full[i] - cur;
      long long new_cur = min((long long)cur, full[i + 1]);
      ende[i] = cur - new_cur;
      cur = new_cur;
    }
    bool success = true;
    int open = 0;
    for (int i = MAX - 1; i >= 0; i--) {
      open += ende[i];
      open -= notfull[i];
      open -= extra[i + 1];
      if (open < 0) {
        success = false;
        break;
      }
    }
    open -= extra[0];
    if (open < 0) {
      success = false;
    }
    if (success) {
      high = mid;
    } else {
      low = mid + 1;
    }
  }
  if (low > full[0]) {
    puts("-1");
    return 0;
  }
  for (int i = low; i <= full[0]; i++) {
    printf("%d ", i);
  }
  puts("");
  return 0;
}
