A10801. Prairie Partition

It can be shown that any positive integer $x$ can be uniquely represented as $x=1+2+4+...+2^{k-1}+r$ , where $k$ and $r$ are integers, $k>=0$ , $0<r<=2^{k}$ . 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!

The first line contains a single integer $n$ ( $1<=n<=10^{5}$ ) — the number of numbers given from Alice to Borys.

The second line contains $n$ integers $a_{1},a_{2},...,a_{n}$ ( $1<=a_{i}<=10^{12}$ ; $a_{1}<=a_{2}<=...<=a_{n}$ ) — the numbers given from Alice to Borys.

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.

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 $[3,3,3]$ .