CF1738D.Permutation Addicts

Given a permutation $a_1, a_2, \dots, a_n$ of integers from $1$ to $n$ , and a threshold $k$ with $0 \leq k \leq n$ , you compute a sequence $b_1, b_2, \dots, b_n$ as follows.

For every $1 \leq i \leq n$ in increasing order, let $x = a_i$ .

• If $x \leq k$ , set $b_{x}$ to the last element $a_j$ ( $1 \leq j < i$ ) that $a_j > k$ . If no such element $a_j$ exists, set $b_{x} = n+1$ .
• If $x > k$ , set $b_{x}$ to the last element $a_j$ ( $1 \leq j < i$ ) that $a_j \leq k$ . If no such element $a_j$ exists, set $b_{x} = 0$ .

Unfortunately, after the sequence $b_1, b_2, \dots, b_n$ has been completely computed, the permutation $a_1, a_2, \dots, a_n$ and the threshold $k$ are discarded.

Now you only have the sequence $b_1, b_2, \dots, b_n$ . Your task is to find any possible permutation $a_1, a_2, \dots, a_n$ and threshold $k$ that produce the sequence $b_1, b_2, \dots, b_n$ . It is guaranteed that there exists at least one pair of permutation $a_1, a_2, \dots, a_n$ and threshold $k$ that produce the sequence $b_1, b_2, \dots, b_n$ .

A permutation of integers from $1$ to $n$ is a sequence of length $n$ which contains all integers from $1$ to $n$ exactly once.

Each test contains multiple test cases. The first line contains an integer $t$ ( $1 \leq t \leq 10^5$ ) — the number of test cases. The following lines contain the description of each test case.

The first line of each test case contains an integer $n$ ( $1 \leq n \leq 10^5$ ), indicating the length of the permutation $a$ .

The second line of each test case contains $n$ integers $b_1, b_2, \dots, b_n$ ( $0 \leq b_i \leq n+1$ ), indicating the elements of the sequence $b$ .

It is guaranteed that there exists at least one pair of permutation $a_1, a_2, \dots, a_n$ and threshold $k$ that produce the sequence $b_1, b_2, \dots, b_n$ .

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

For each test case, output the threshold $k$ ( $0 \leq k \leq n$ ) in the first line, and then output the permutation $a_1, a_2, \dots, a_n$ ( $1 \leq a_i \leq n$ ) in the second line such that the permutation $a_1, a_2, \dots, a_n$ and threshold $k$ produce the sequence $b_1, b_2, \dots, b_n$ . If there are multiple solutions, you can output any of them.

For the first test case, permutation $a = [1,3,2,4]$ and threshold $k = 2$ will produce sequence $b$ as follows.

• When $i = 1$ , $x = a_i = 1 \leq k$ , there is no $a_j$ ( $1 \leq j < i$ ) that $a_j > k$ . Therefore, $b_1 = n + 1 = 5$ .
• When $i = 2$ , $x = a_i = 3 > k$ , the last element $a_j$ that $a_j \leq k$ is $a_1$ . Therefore, $b_3 = a_1 = 1$ .
• When $i = 3$ , $x = a_i = 2 \leq k$ , the last element $a_j$ that $a_j > k$ is $a_2$ . Therefore, $b_2 = a_2 = 3$ .
• When $i = 4$ , $x = a_i = 4 > k$ , the last element $a_j$ that $a_j \leq k$ is $a_3$ . Therefore, $b_4 = a_3 = 2$ .

Finally, we obtain sequence $b = [5,3,1,2]$ . For the second test case, permutation $a = [1,2,3,4,5,6]$ and threshold $k = 3$ will produce sequence $b$ as follows.

• When $i = 1, 2, 3$ , $a_i \leq k$ , there is no $a_j$ ( $1 \leq j < i$ ) that $a_j > k$ . Therefore, $b_1 = b_2 = b_3 = n + 1 = 7$ .
• When $i = 4, 5, 6$ , $a_i > k$ , the last element $a_j$ that $a_j \leq k$ is $a_3$ . Therefore, $b_4 = b_5 = b_6 = a_3 = 3$ .

Finally, we obtain sequence $b = [7,7,7,3,3,3]$ . For the third test case, permutation $a = [6,5,4,3,2,1]$ and threshold $k = 3$ will produce sequence $b$ as follows.

• When $i = 1, 2, 3$ , $a_i > k$ , there is no $a_j$ ( $1 \leq j < i$ ) that $a_j \leq k$ . Therefore, $b_4 = b_5 = b_6 = 0$ .
• When $i = 4, 5, 6$ , $a_i \leq k$ , the last element $a_j$ that $a_j > k$ is $a_3$ . Therefore, $b_1 = b_2 = b_3 = a_3 = 4$ .

Finally, we obtain sequence $b = [4,4,4,0,0,0]$ .