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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

class Graph<T> {

	private HashMap<T, Set<T> > graph;

	public Graph() {
		this.graph = new HashMap<T, Set<T> >();
	}

	public void addNode(T node) {
		if (! this.graph.containsKey(node)) {
			this.graph.put(node, new HashSet<T>());
		}
	}

	public void addEdge(T src, T dest) {
		this.graph.get(src).add(dest);
	}

	public Set<T> getEdges(T node) {
		return this.graph.get(node);
	}

	public Set<T> getNodes() {
		return this.graph.keySet();
	}

	private boolean has_cycle(T node, Set<T> active, Set<T> seen) {

		if (seen.contains(node)) {
			return false;
		}
		seen.add(node);
		active.add(node);
		for (T next_node : this.getEdges(node)) {
			if (active.contains(next_node) || this.has_cycle(next_node, active, seen)) {
				return true;
			}
		}
		active.remove(node);
		return false;
	}

	public Set<T> find_cycle(T start_node) {

		Set<T> seen = new HashSet<T>();
		Set<T> active = new HashSet<T>();

		this.has_cycle(start_node, active, seen);
		return active;

	}

	/*
	static class Node {

		int num;


		Node(int num) {
			this.num = num;
		}

	}

	static class Edge {

		Node src;
		Node dest;

		Edge(Node src, Node dest) {
			this.src = src;
			this.dest = dest;
		}

	}
	*/


}

public class Main {


	public static void main(String[] args) {

		Scanner scan = new Scanner(System.in);
		Graph<Integer> graph = new Graph<Integer>();

		int v = scan.nextInt();
		int e = scan.nextInt();
		for (int i = 0; i < e; i++) {
			int s = scan.nextInt();
			int t = scan.nextInt();

			graph.addNode(s);
			graph.addNode(t);
			graph.addEdge(s, t);
		}

		for (int n : graph.getNodes()) {
			if (graph.find_cycle(n).size() > 0) {
				Set<Integer> a = graph.find_cycle(n);
				System.out.println(a.size());
				for(int b:a) {
					System.out.println(b);
				}
				return;
			}
		}
		System.out.println("-1");
	}

}