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)

import java.io._
import java.util.StringTokenizer
import scala.util.control.Breaks.{breakable,break}
import scala.collection.mutable.{Map, Queue, PriorityQueue}

object Main extends App {
  val in = new InputReader(System.in)
  val pw = new PrintWriter(System.out)

  val N, M = in.next().toInt
  val graph = Array.fill[List[Int]](N+1)(Nil)
  for (_ <- 1 to M) {
    val a, b = in.next().toInt
    graph(a) = b :: graph(a)
  }

  def mkAns(now: Int, lst: List[Int]): List[Int] = {
    lst match {
      case a :: rest =>
        if (a == now)
          List(a)
        else
          a :: mkAns(now, rest)
    }
  }

  //
  val done = new Array[Boolean](N+1)
  var ans: List[Int] = Nil
  breakable {
    for (i <- 1 to N) {
      if (!done(i)) {
        done(i) = true

        val q = new Queue[(Int, List[Int])]
        //val dp = new Array[Boolean](N+1)
        //dp(i) = true
        q.enqueue((i, List()))

        breakable {
          while (!q.isEmpty) {
            val (now, alr) = q.dequeue()
            if (!done(now))
              done(now) = true

            if (alr.contains(now)) {
              ans = mkAns(now, alr)
            }
            else {
              for (e <- graph(now))
                q.enqueue((e, now :: alr))
            }
          }
        }
      }

      if (ans != Nil)
        break
    }
  }

  if (ans == Nil)
    pw.println(-1)
  else {
    pw.println(ans.length)
    ans.foreach { x => pw.println(x) }
  }
  pw.flush()
}


class InputReader(stream: InputStream) {
  private val reader = new BufferedReader(new InputStreamReader(stream))
  private var tokenizer: StringTokenizer = new StringTokenizer(reader.readLine())
  def next(): String = {
    while (!tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine())
    }
    tokenizer.nextToken()
  }
}