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 a[100010];
int power[100], between[100];
bool judge(int x) {
  int r = 0;
  for (int i = 0; i < 63; i++) {
    if (i > 0) r += between[i - 1];
    if (power[i] > x) {
      r += power[i] - x;
    } else if (power[i] < x) {
      r -= min(r, x - power[i]);
      x = power[i];
    }
  }
  return r <= 0;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  long long now = 1;
  int idx = 0;
  for (int i = 0; i < n; i++) {
    while (a[i] > now) {
      idx++;
      now *= 2;
    }
    if (a[i] == now) {
      power[idx]++;
    }
    if (a[i] < now) {
      between[idx - 1]++;
    }
  }
  int y = power[0] + 1, x = 1;
  while (x < y) {
    int mid = x + (y - x) / 2;
    if (judge(mid))
      y = mid;
    else
      x = mid + 1;
  }
  if (x > power[0]) {
    cout << -1 << endl;
  } else {
    for (int i = x; i <= power[0]; i++) {
      if (i > x) putchar(' ');
      printf("%d", i);
    }
    puts("");
  }
  return 0;
}
