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 = 100000;
int n, cnt;
long long rec[MaxN + 5];
long long a[MaxN + 5];
long long fac[MaxN + 5];
int temp[MaxN + 5];
bool flag[MaxN + 5];
void Init() {
  long long now = 0;
  cnt = 0;
  memset(flag, 0, sizeof(flag));
  memset(fac, 0, sizeof(fac));
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    if (a[i] == 1) cnt++;
    while (a[i] >= (1LL << (now + 1))) now++;
    if (a[i] == (1LL << now)) flag[i] = true;
    fac[i] = now;
  }
}
bool Check(int mid) {
  int useless = 0;
  int c[100];
  memset(c, 0, sizeof(c));
  for (int i = 1; i <= mid; i++) c[0]++;
  for (int i = mid + 1; i <= cnt; i++) rec[++useless] = fac[i];
  for (int i = cnt + 1; i <= n; i++) {
    if (!flag[i])
      rec[++useless] = fac[i];
    else {
      if (c[fac[i] - 1] > 0) {
        c[fac[i] - 1]--;
        c[fac[i]]++;
      } else
        rec[++useless] = fac[i];
    }
  }
  int tot = 0;
  for (int i = 0; i <= 63; i++)
    while (c[i] > 0) temp[++tot] = i, c[i]--;
  for (int i = useless; i >= 1; i--) {
    if (tot == 0) return false;
    if (temp[tot] < rec[i]) return false;
    tot--;
  }
  return true;
}
void Solve() {
  int l = 0, r = cnt + 1;
  while (l < r - 1) {
    int mid = (l + r) >> 1;
    if (Check(mid))
      r = mid;
    else
      l = mid;
  }
  if (r > cnt)
    printf("-1");
  else
    for (int i = r; i <= cnt; i++) printf("%d ", i);
  printf("\n");
}
int main() {
  while (~scanf("%d", &n)) {
    Init();
    Solve();
  }
  return 0;
}
