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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;

class F
{
	static void Main()
	{
		Func<int[]> read = () => Console.ReadLine().Split().Select(int.Parse).ToArray();
		var h = read();
		var n = h[0];
		var rs = new int[h[1]].Select(_ => read()).ToArray();
		var map = rs.GroupBy(x => x[0]).ToDictionary(g => g.Key, g => g.Select(x => x[1]).ToArray());

		var u = new bool[n + 1];
		var t = new bool[n + 1];
		var q = new Stack<int>();
		var qi = new int[n + 1];
		var ps = rs.Select(x => x[1]).Distinct().ToArray();
		foreach (var sp in ps)
		{
			if (u[sp]) continue;
			u[sp] = t[sp] = true;
			q.Push(sp);
			while (q.Any())
			{
				var p = q.Pop();
				if (!map.ContainsKey(p)) continue;
				foreach (var np in map[p])
				{
					if (u[np])
					{
						if (t[np])
						{
							var closed = true;
							var l = new List<int> { p };
							var xp = p;
							while (xp != np)
							{
								if (qi[xp] == 0) { closed = false; break; }
								l.Add(xp = qi[xp]);
							}
							if (!closed) continue;

							var cs = new HashSet<int>(l);
							if (rs.Count(x => cs.Contains(x[0]) && cs.Contains(x[1])) == cs.Count)
							{
								Console.WriteLine(cs.Count);
								Console.WriteLine(string.Join("\n", cs));
								return;
							}
							else continue;
						}
						else
						{
							t[np] = true;
							continue;
						}
					}

					u[np] = t[np] = true;
					qi[np] = p;
					q.Push(np);
				}
			}
			Array.Clear(t, 0, n + 1);
		}
		Console.WriteLine(-1);
	}
}