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 = 2e3 + 20;
long long n, a[N];
int power[N];
int between[N];
bool check(int x) {
  int t = power[0] - x;
  for (int i = 1; i <= 63; i++) {
    if (x >= power[i]) {
      t -= min(x - power[i], t);
      x = power[i];
    } else
      t += power[i] - x;
    t += between[i];
  }
  return t <= x;
}
int main() {
  while (cin >> n) {
    long long x;
    memset(power, 0, sizeof(power));
    memset(between, 0, sizeof(between));
    for (int i = 0; i < n; i++) {
      scanf("%I64d", &x);
      int cc = 0;
      bool good = true;
      while (x > 1) {
        if (x & 1) good = false;
        cc++;
        x >>= 1;
      }
      if (good)
        power[cc]++;
      else
        cc++, between[cc - 1]++;
    }
    int ans = -1, l = 1, r = power[0];
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (check(mid)) {
        ans = mid;
        r = mid - 1;
      } else
        l = mid + 1;
    }
    if (ans == -1)
      cout << ans;
    else
      for (int i = ans; i <= power[0]; i++) cout << i << ' ';
    cout << endl;
  }
  return 0;
}
