A16040. Insert Zero and Invert Prefix

You have a sequence $a_1, a_2, \ldots, a_n$ of length $n$ , each element of which is either $0$ or $1$ , and a sequence $b$ , which is initially empty.

You are going to perform $n$ operations. On each of them you will increase the length of $b$ by $1$ .

- On the $i$ -th operation you choose an integer $p$ between $0$ and $i-1$ . You insert $0$ in the sequence $b$ on position $p+1$ (after the first $p$ elements), and then you invert the first $p$ elements of $b$ .
- More formally: let's denote the sequence $b$ before the $i$ -th ( $1 \le i \le n$ ) operation as $b_1, b_2, \ldots, b_{i-1}$ . On the $i$ -th operation you choose an integer $p$ between $0$ and $i-1$ and replace $b$ with $\overline{b_1}, \overline{b_2}, \ldots, \overline{b_{p}}, 0, b_{p+1}, b_{p+2}, \ldots, b_{i-1}$ . Here, $\overline{x}$ denotes the binary inversion. Hence, $\overline{0} = 1$ and $\overline{1} = 0$ .

You can find examples of operations in the Notes section.

Determine if there exists a sequence of operations that makes $b$ equal to $a$ . If such sequence of operations exists, find it.

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.

The first line of each test case contains one integer $n$ ( $1 \le n \le 10^5$ ) — the length of the sequence $a$ .

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ( $0 \le a_i \le 1$ ) — the sequence $a$ .

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$ .

For each test case:

- output "NO", if it is impossible to make $b$ equal to $a$ using the given operations;
- otherwise, output "YES" in the first line and $n$ integers $p_1, p_2, \ldots, p_n$ ( $0 \le p_i \le i-1$ ) in the second line — the description of sequence of operations that makes $b$ equal to $a$ . Here, $p_i$ should be the integer you choose on the $i$ -th operation. If there are multiple solutions, you can output any of them.

In the first test case,

1. Before the first operation, $b = [\,]$ . You choose $p = 0$ and replace $b$ with $[\, \underline{0} \,]$
2. On the second operation you choose $p = 0$ and replace $b$ with $[\, \underline{0}, 0 \,]$ .
3. On the third operation you choose $p = 2$ and replace $b$ with $[\, 1, 1, \underline{0} \,]$ .
4. On the fourth operation you choose $p = 1$ and replace $b$ with $[\, 0, \underline{0}, 1, 0 \,]$ .
5. On the fifth operation you choose $p = 3$ and replace $b$ with $[\, 1, 1, 0, \underline{0}, 0 \,]$ .

Hence, sequence $b$ changes in the following way: $[\,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0} \,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0}, 0 \,]$ $\xrightarrow{p \, = \, 2}$ $[\, 1, 1, \underline{0} \,]$ $\xrightarrow{p \, = \, 1}$ $[\, 0, \underline{0}, 1, 0 \,]$ $\xrightarrow{p \, = \, 3}$ $[\, 1, 1, 0, \underline{0}, 0 \,]$ . In the end the sequence $b$ is equal to the sequence $a$ , so this way to perform operations is one of the correct answers.

In the second test case, $n = 1$ and the only achiveable sequence $b$ is $[\, 0 \, ]$ .

In the third test case, there are six possible sequences of operations:

1. $[\,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0} \,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0}, 0 \,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0}, 0, 0 \,]$ .
2. $[\,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0} \,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0}, 0 \,]$ $\xrightarrow{p \, = \, 1}$ $[\, 1, \underline{0}, 0 \,]$ .
3. $[\,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0} \,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0}, 0 \,]$ $\xrightarrow{p \, = \, 2}$ $[\, 1, 1, \underline{0} \,]$ .
4. $[\,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0} \,]$ $\xrightarrow{p \, = \, 1}$ $[\, 1, \underline{0} \,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0}, 1, 0 \,]$ .
5. $[\,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0} \,]$ $\xrightarrow{p \, = \, 1}$ $[\, 1, \underline{0} \,]$ $\xrightarrow{p \, = \, 1}$ $[\, 0, \underline{0}, 0 \,]$ .
6. $[\,]$ $\xrightarrow{p \, = \, 0}$ $[\, \underline{0} \,]$ $\xrightarrow{p \, = \, 1}$ $[\, 1, \underline{0} \,]$ $\xrightarrow{p \, = \, 2}$ $[\, 0, 1, \underline{0} \,]$ .

None of them makes $b$ equal to $[\, 0, 1, 1 \,]$ , so the answer is "NO".