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;
template <typename T>
bool NextInteger(T &res) {
  char c;
  bool negative = false;
  while (c = getchar(), c != '-' && !('0' <= c && c <= '9'))
    if (c == EOF) return false;
  if (c == '-') negative = true, c = getchar();
  res = 0;
  while ('0' <= c && c <= '9') res *= 10, res += c - '0', c = getchar();
  if (negative) res *= -1;
  return true;
}
template <typename T>
void PrintInteger(T x) {
  if (x < 0) putchar('-'), x *= -1;
  if (x > 9) PrintInteger(x / 10);
  putchar(x % 10 + '0');
}
array<long long, maxn + 1> a, b, c;
int n;
int check(int x) {
  int cnt_b = 0, cur_n = 0;
  long long cur_m = 1;
  for (int i = 0; i < x; i++) c[i] = 0;
  for (int i = 0; i < n; i++) {
    if (cur_m < a[i]) cur_n = 0, cur_m *= 2;
    if (a[i] == cur_m && a[i] / 2 == c[cur_n])
      c[cur_n++] = a[i];
    else
      b[cnt_b++] = a[i];
    if (cur_n == x || c[cur_n] != cur_m / 2) cur_n = 0, cur_m *= 2;
  }
  if (c[x - 1] == 0 || cnt_b > x) return false;
  for (int i = 0; i < cnt_b; i++) {
    if (b[cnt_b - 1 - i] > 2 * c[i] - 1) return false;
  }
  return true;
}
int main() {
  NextInteger(n);
  for (int i = 0; i < n; i++) NextInteger(a[i]);
  int cnt = upper_bound(a.begin(), a.begin() + n, 1) - a.begin() + 1;
  int l = 1, r = cnt;
  while (l < r) {
    int mid = l + (r - l) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  if (l == cnt)
    printf("-1\n");
  else {
    for (int i = l; i < cnt; i++) printf("%d ", i);
    printf("\n");
  }
  return 0;
}
