CF1810B.Candies

This problem is about candy. Initially, you only have $1$ candy, and you want to have exactly $n$ candies.

You can use the two following spells in any order at most $40$ times in total.

• Assume you have $x$ candies now. If you use the first spell, then $x$ candies become $2x-1$ candies.
• Assume you have $x$ candies now. If you use the second spell, then $x$ candies become $2x+1$ candies.

Construct a sequence of spells, such that after using them in order, you will have exactly $n$ candies, or determine it's impossible.

Each test contains multiple test cases. The first line contains a single integer $t$ ( $1 \le t \le 10^4$ ) — the number of test cases. Their description follows.

Each test case contains one line with a single integer $n$ ( $2 \le n \le 10^9$ ) — the required final number of candies.

For each test case, output the following.

If it's possible to eventually have $n$ candies within $40$ spells, in the first line print an integer $m$ ( $1 \le m \le 40$ ), representing the total number of spells you use.

In the second print $m$ integers $a_{1}, a_{2}, \ldots, a_{m}$ ( $a_{i}$ is $1$ or $2$ ) separated by spaces, where $a_{i} = 1$ means that you use the first spell in the $i$ -th step, while $a_{i} = 2$ means that you use the second spell in the $i$ -th step.

Note that you do not have to minimize $m$ , and if there are multiple solutions, you may output any one of them.

If it's impossible, output $-1$ in one line.

For $n=3$ , you can just use the second spell once, and then have $2 \cdot 1 + 1 = 3$ candies.

For $n=7$ , you can use the second spell twice. After the first step, you will have $3$ candies. And after the second step, you will have $2 \cdot 3 + 1 = 7$ candies.