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 = 100 + 5;
int n;
long long m;
int a[N], b[N];
inline bool judge(int x) {
  int invalid = a[0] - x + b[0];
  for (int i = 1; i < 60; i++) {
    if (a[i] <= x) {
      invalid -= min(invalid, x - a[i]);
      x = a[i];
    } else
      invalid += a[i] - x;
    invalid += b[i];
  }
  return invalid <= x;
}
int main() {
  scanf("%d", &n);
  while (n--) {
    scanf("%lld", &m);
    int bit = 0;
    bool flag = true;
    while (m > 1) {
      if (m & 1) flag = false;
      m >>= 1;
      ++bit;
    }
    if (!flag)
      ++b[bit];
    else
      ++a[bit];
  }
  int L = a[0] / 2, R = a[0];
  int ans = R;
  while (L <= R) {
    int M = L + R >> 1;
    if (judge(M)) {
      ans = M;
      R = M - 1;
    } else {
      L = M + 1;
    }
  }
  if (!judge(ans)) return puts("-1"), 0;
  for (int i = ans; i <= a[0]; i++) printf("%d%c", i, " \n"[i == a[0]]);
  return 0;
}
