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 INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 1e9 + 7;
const long long maxn = 1e6 + 7;
const double eps = 0.00000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
int i, j;
long long k;
int sum;
int a[64], b[64];
int num[64];
void pushin(long long x) {
  int i;
  for (i = 0; i <= 60; i++) {
    if (x == 1ll << i) {
      a[i]++;
      return;
    }
    if (x < 1ll << i) {
      b[i - 1]++;
      return;
    }
  }
}
bool judge(int x) {
  int i, count = 0, prex = x, use;
  for (i = 0; i <= 60; i++) {
    if (a[i] < x) x = a[i];
    num[i] = x;
  }
  for (i = 60; i >= 0; i--) {
    count += a[i] + b[i] - num[i];
    if (count > num[i]) return 0;
  }
  return 1;
}
bool mark;
int n;
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%I64d", &k);
    pushin(k);
  }
  for (i = 1; i <= a[0]; i++)
    if (judge(i)) printf("%d ", i), mark = 1;
  if (!mark) puts("-1");
}
