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;
long long a[64], b[64], c[64];
long long aa[64], bb[64];
vector<long long> ans;
int main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    long long x;
    scanf("%lld", &x);
    if (x == 1) {
      a[0]++;
      continue;
    }
    long long y = 1;
    for (long long j = 1; j <= 64; j++) {
      y = y * 2;
      if (y == x) {
        a[j]++;
        break;
      }
      if (y > x) {
        b[j]++;
        break;
      }
    }
  }
  for (long long i = 1; i <= a[0]; i++) {
    for (long long j = 0; j < 64; j++) aa[j] = a[j], bb[j] = b[j], c[j] = 0;
    long long cnt = i;
    for (long long j = 0; cnt != 0; j++) {
      if (aa[j] >= cnt) {
        aa[j] -= cnt;
      } else {
        c[j] = cnt - aa[j];
        cnt -= (cnt - aa[j]);
        aa[j] = 0;
      }
    }
    int q = 63;
    int flag = 0;
    for (int j = 63; j >= 0; j--) {
      while ((aa[j] + bb[j])) {
        if ((aa[j] + bb[j]) > c[q]) {
          aa[j] -= c[q];
          c[q] = 0;
          q--;
          if (q < j) break;
        } else {
          c[q] -= aa[j] + bb[j];
          aa[j] = bb[j] = 0;
        }
      }
    }
    for (int j = 0; j < 64; j++)
      if ((aa[j] + bb[j]) != 0) flag = 1;
    if (flag == 0) ans.push_back(i);
  }
  if (ans.size() == 0) {
    printf("-1\n");
  }
  for (long long i = 0; i < ans.size(); i++) {
    printf("%lld ", ans[i]);
  }
}
