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;
int n;
long long a[100005];
int check(int x) {
  int i, j, res, ress, t, sum;
  long long A, AA;
  i = 0;
  ress = 0;
  for (j = 0; j < 60; j++) {
    A = (1LL << j);
    res = 0;
    for (; i < n; i++) {
      if (a[i] == A) {
        res++;
      } else if (a[i] < A) {
        ress++;
      } else
        break;
    }
    if (ress > x) return 0;
    if (x > res) {
      ress -= (x - res);
      if (ress < 0) ress = 0;
      x = res;
    }
    if (res > x) ress += (res - x);
  }
  return 1;
}
int main(void) {
  int i, j, L, R, M, ans;
  scanf("%d", &n);
  L = 0;
  R = 0;
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    if (a[i] == 1) R++;
  }
  sort(a, a + n);
  ans = R;
  while (L < R) {
    M = (L + R) / 2;
    if (check(M))
      R = M;
    else
      L = M + 1;
  }
  if (L == 0 || check(L) == 0) {
    printf("-1\n");
    return 0;
  }
  for (i = L; i <= ans; i++) {
    printf("%d ", i);
  }
  printf("\n");
}
