Problem:
Score : 600 points

Problem Statement

Given is a directed graph G with N vertices and M edges.

The vertices are numbered 1 to N, and the i-th edge is directed from Vertex A_i to Vertex B_i.

It is guaranteed that the graph contains no self-loops or multiple edges.

Determine whether there exists an induced subgraph (see Notes) of G such that the in-degree and out-degree of every vertex are both 1. If the answer is yes, show one such subgraph.

Here the null graph is not considered as a subgraph.

Notes

For a directed graph G = (V, E), we call a directed graph G' = (V', E') satisfying the following conditions an induced subgraph of G:

V' is a (non-empty) subset of V.

E' is the set of all the edges in E that have both endpoints in V'.

Constraints

1 \leq N \leq 1000

0 \leq M \leq 2000

1 \leq A_i,B_i \leq N

A_i \neq B_i

All pairs (A_i, B_i) are distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
:
A_M B_M

Output

If there is no induced subgraph of G that satisfies the condition, print -1.
Otherwise, print an induced subgraph of G that satisfies the condition, in the following format:

K
v_1
v_2
:
v_K

This represents the induced subgraph of G with K vertices whose vertex set is \{v_1, v_2, \ldots, v_K\}. (The order of v_1, v_2, \ldots, v_K does not matter.)
If there are multiple subgraphs of G that satisfy the condition, printing any of them is accepted.

Sample Input 1

4 5
1 2
2 3
2 4
4 1
4 3

Sample Output 1

3
1
2
4

The induced subgraph of G whose vertex set is \{1, 2, 4\} has the edge set \{(1, 2), (2, 4), (4, 1)\}. The in-degree and out-degree of every vertex in this graph are both 1.

Sample Input 2

4 5
1 2
2 3
2 4
1 4
4 3

Sample Output 2

-1

There is no induced subgraph of G that satisfies the condition.

Sample Input 3

6 9
1 2
2 3
3 4
4 5
5 6
5 1
5 2
6 1
6 2

Sample Output 3

4
2
3
4
5

Solution in Scala (2.11.7) (judge verdict: Wrong Answer)

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)

    val V, E = sc.nextInt()
    val G = Array.fill(V + 1)(List[Int]())
    (0 until E) foreach { _ =>
      val u, v = sc.nextInt()
      G(u) ::= v
    }

    def solve(V: Int, E: Int, G: Array[List[Int]]): Option[Set[Int]] = {
      val visited = new Array[Boolean](V+1)
      val cycle = new Array[Boolean](V+1)

      def findCyclic(stack: List[Int]): Boolean =
        if (stack.nonEmpty) {
          val u = stack.head
          if (u > 0) {
            if (cycle(u)) {
              skipWhileStartNode(u, stack.tail)
              true
            } else {
              cycle(u) = true
              visited(u) = true
              findCyclic(G(u).foldRight(-u :: stack.tail)((v, stack) => v :: stack))
            }
          } else {
            cycle(-u) = false
            findCyclic(stack.tail)
          }
        } else false

      def skipWhileStartNode(start: Int, stack: List[Int]): Unit =
        if (stack.nonEmpty) {
          val u = stack.head
          if (-u == start) removeOthers(stack.tail)
          else skipWhileStartNode(start, stack.tail)
        }

      def removeOthers(stack: List[Int]): Unit =
        if (stack.nonEmpty) {
          val u = stack.head
          if (u < 0) cycle(-u) = false
          removeOthers(stack.tail)
        }

      def loop(s: Int): Option[Set[Int]] =
        if (s <= V) {
          if (!visited(s) && findCyclic(s :: Nil)) Some((1 to V) filter cycle toSet)
          else loop(s + 1)
        } else None

      loop(1)
    }

    solve(V, E, G) match {
      case Some(set) => println(set.size); println(set.mkString("\n"))
      case None => println(-1)
    }
  }

}