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 maxn = 1e5 + 10, LOG = 70;
int n, power[LOG], between[LOG], extra[LOG], ende[LOG];
bool check(int x) {
  int cur = x;
  for (int i = 0; i < LOG; i++) {
    extra[i] = power[i] - cur;
    int newCur = min(cur, power[i + 1]);
    ende[i] = cur - newCur;
    cur = newCur;
  }
  int res = 0;
  for (int i = LOG - 1; i >= 0; i--) {
    res += ende[i];
    res -= between[i];
    res -= extra[i + 1];
    if (res < 0) return false;
  }
  return res - extra[0] >= 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    long long x;
    scanf("%lld", &x);
    int cur = 0;
    bool good = true;
    while (x > 1) {
      if (x & 1) good = false;
      x >>= 1;
      cur++;
    }
    if (good)
      power[cur]++;
    else
      between[cur]++;
  }
  int l = 1, r = power[0];
  while (l <= r) {
    int m = (l + r) >> 1;
    if (check(m))
      r = m - 1;
    else
      l = m + 1;
  }
  if (l <= power[0] && check(l)) {
    for (int i = l; i <= power[0]; i++) printf("%d ", i);
  } else {
    printf("-1\n");
  }
  return 0;
}
