CF1787F.Inverse Transformation

A permutation scientist is studying a self-transforming permutation $a$ consisting of $n$ elements $a_1,a_2,\ldots,a_n$ .

A permutation is a sequence of integers from $1$ to $n$ of length $n$ containing each number exactly once. For example, $[1]$ , $[4, 3, 5, 1, 2]$ are permutations, while $[1, 1]$ , $[4, 3, 1]$ are not.

The permutation transforms day by day. On each day, each element $x$ becomes $a_x$ , that is, $a_x$ becomes $a_{a_x}$ . Specifically:

• on the first day, the permutation becomes $b$ , where $b_x = a_{a_x}$ ;
• on the second day, the permutation becomes $c$ , where $c_x = b_{b_x}$ ;
• $\ldots$

For example, consider permutation $a = [2,3,1]$ . On the first day, it becomes $[3,1,2]$ . On the second day, it becomes $[2,3,1]$ .You're given the permutation $a'$ on the $k$ -th day.

Define $\sigma(x) = a_x$ , and define $f(x)$ as the minimal positive integer $m$ such that $\sigma^m(x) = x$ , where $\sigma^m(x)$ denotes $\underbrace{\sigma(\sigma(\ldots \sigma}_{m \text{ times}}(x) \ldots))$ .

For example, if $a = [2,3,1]$ , then $\sigma(1) = 2$ , $\sigma^2(1) = \sigma(\sigma(1)) = \sigma(2) = 3$ , $\sigma^3(1) = \sigma(\sigma(\sigma(1))) = \sigma(3) = 1$ , so $f(1) = 3$ . And if $a=[4,2,1,3]$ , $\sigma(2) = 2$ so $f(2) = 1$ ; $\sigma(3) = 1$ , $\sigma^2(3) = 4$ , $\sigma^3(3) = 3$ so $f(3) = 3$ .

Find the initial permutation $a$ such that $\sum\limits^n_{i=1} \dfrac{1}{f(i)}$ is minimum possible.

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

The first line of each test case contains two integers $n$ and $k$ ( $1 \le n \le 2 \cdot 10^5$ ; $1 \le k \le 10^9$ ) — the length of $a$ , and the last day.

The second line contains $n$ integers $a'_1,a'_2,\ldots,a'_n$ ( $1 \le a'_i \le n$ ) — the permutation on the $k$ -th day.

It's guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ .

For each test case, if at least one initial $a$ consistent with the given $a'$ exists, print "YES", then print $n$ integers $a_1,a_2,\ldots,a_n$ — the initial permutation with the smallest sum $\sum\limits^n_{i=1} \dfrac{1}{f(i)}$ . If there are multiple answers with the smallest sum, print any.

If there are no valid permutations, print "NO".

In the second test case, the initial permutation can be $a = [6,2,5,7,1,3,4]$ , which becomes $[3,2,1,4,6,5,7]$ on the first day and $a' = [1,2,3,4,5,6,7]$ on the second day (the $k$ -th day). Also, among all the permutations satisfying that, it has the minimum $\sum\limits^n_{i=1} \dfrac{1}{f(i)}$ , which is $\dfrac{1}{4}+\dfrac{1}{1}+\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{2}=3$ .

In the fifth test case, the initial permutation can be $a = [4,2,1,3]$ , which becomes $[3,2,4,1]$ on the first day, $[4,2,1,3]$ on the second day, and so on. So it finally becomes $a' = [4,2,1,3]$ on the $8$ -th day (the $k$ -th day). And it has the minimum $\sum\limits^n_{i=1} \dfrac{1}{f(i)} = \dfrac{1}{3} + \dfrac{1}{1} + \dfrac{1}{3} + \dfrac{1}{3} = 2$ .