There is a directed graph with $N$ vertices and $N$ edges.  
The $i$-th edge goes from vertex $i$ to vertex $A_i$. (The constraints guarantee that $i \neq A_i$.)  
Find a directed cycle without the same vertex appearing multiple times.  
It can be shown that a solution exists under the constraints of this problem.

 Notes

The sequence of vertices $B = (B_1, B_2, \dots, B_M)$ is called a directed cycle when all of the following conditions are satisfied:

-   $M \geq 2$
-   The edge from vertex $B_i$ to vertex $B_{i+1}$ exists. $(1 \leq i \leq M-1)$
-   The edge from vertex $B_M$ to vertex $B_1$ exists.
-   If $i \neq j$, then $B_i \neq B_j$.

The input is given from Standard Input in the following format:

```
$N$
$A_1$ $A_2$ $\dots$ $A_N$
```

Print a solution in the following format:

```
$M$
$B_1$ $B_2$ $\dots$ $B_M$
```

$M$ is the number of vertices, and $B_i$ is the $i$-th vertex in the directed cycle.  
The following conditions must be satisfied:

-   $2 \le M$
-   $B_{i+1} = A_{B_i}$ ( $1 \le i \le M-1$ )
-   $B_{1} = A_{B_M}$
-   $B_i \neq B_j$ ( $i \neq j$ )

If multiple solutions exist, any of them will be accepted.

-   All input values are integers.
-   $2 \le N \le 2 \times 10^5$
-   $1 \le A_i \le N$
-   $A_i \neq i$

4
7 5 3 2

7→5→3→2→7 is indeed a directed cycle.

Here is the graph corresponding to this input:

Here are other acceptable outputs:

4
2 7 5 3

3
4 1 6

Note that the graph may not be connected.

2
1 2

This case contains both of the edges 1→2 and 2→1.
In this case, 1→2→1 is indeed a directed cycle.

Here is the graph corresponding to this input, where 1↔2 represents the existence of both 1→2 and 2→1:

3
2 7 8

Here is the graph corresponding to this input: