807_E. Prairie Partition
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].
```python
from collections import Counter
from math import log2

powers = [2 ** p for p in range(64)]


def can():
    near = Counter()
    for num in set(nums):
        p = int(log2(num))
        near[p] += nums[num]
    seq_cp = seqs - Counter()
    for num in set(near):
        cnt = near[num]
        while cnt != 0 and num < 64:
            dif = min(cnt, seq_cp[powers[num]])
            cnt -= dif
            seq_cp[powers[num]] -= dif
            num += 1
        if cnt > 0:
            return False
    return True


n = int(input())

seq = list(map(int, input().split()))
cnt = Counter(seq)
nums = Counter(seq)

seqs = Counter()
cur_cnt = cnt[1]
cur_pow = 1
while cur_cnt != 0:
    nums[cur_pow] -= cur_cnt
    cur_pow *= 2
    if cur_cnt > cnt[cur_pow]:
        seqs[cur_pow // 2] = cur_cnt - cnt[cur_pow]
        cur_cnt = cnt[cur_pow]

for num in set(nums):
    addition = nums[num]
    nums[num] = 0
    nums[2 ** int(log2(num))] += addition

# remove elements with zero count
nums -= Counter()
seqs -= Counter()

cur_len = sum(seqs[num] for num in set(seqs))
res = []
cur_pow = 1

while can():
    res.append(cur_len)
    while seqs[cur_pow] == 0 and cur_pow <= 2 ** 63:
        cur_pow *= 2
    if cur_pow > 2 ** 63:
        break
    other_pow = 1
    while other_pow <= cur_pow:
        nums[other_pow] += 1
        other_pow *= 2
    seqs[cur_pow] -= 1
    cur_len -= 1

print(-1 if len(res) == 0 else ' '.join(map(str, reversed(res))))

```