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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, cnt[45], mxx[200010], cntt[45];
long long ote[200010], tbl[45], tmp[200010];
inline int check(int x) {
  memset(mxx, 0, sizeof mxx);
  for (int i = 0; i <= 40; i++) cntt[i] = cnt[i];
  int mx = x;
  for (int i = 0; i <= ote[0]; i++) tmp[i] = ote[i];
  for (int i = 0; i <= 40 && mx; i++) {
    int s = 1;
    while (cntt[i] && s <= mx) mxx[s++]++, cntt[i]--;
    mx = s - 1;
    while (cntt[i]) cntt[i]--, tmp[++tmp[0]] = tbl[i];
  }
  sort(tmp + 1, tmp + 1 + tmp[0]);
  if (tmp[0] > x) return 0;
  for (int i = 1; i <= tmp[0]; i++)
    if (tmp[tmp[0] - i + 1] >= tbl[mxx[i]]) return 0;
  return 1;
}
int main() {
  tbl[0] = 1;
  for (int i = 1; i <= 40; i++) tbl[i] = tbl[i - 1] << 1;
  n = read();
  for (int i = 1; i <= n; i++) {
    long long xx = read(), fg = 0;
    for (int j = 0; j <= 40; j++)
      if (tbl[j] == xx) cnt[j]++, fg = 1;
    if (!fg) ote[++ote[0]] = xx;
  }
  sort(ote + 1, ote + 1 + ote[0]);
  int L = 1, R = cnt[0], Ans = cnt[0] + 1;
  while (L <= R) {
    int mid = (L + R) >> 1;
    if (check(mid))
      R = mid - 1, Ans = mid;
    else
      L = mid + 1;
  }
  if (Ans == cnt[0] + 1)
    puts("-1");
  else {
    for (int i = Ans; i <= cnt[0]; i++) printf("%d ", i);
    puts("");
  }
  return 0;
}
