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.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class Main {
    private static Graph g;

    private static int[] color;
    private static LinkedList<Integer> q = new LinkedList<>();

    static void dfs(int v) {
        color[v] = 1;
        q.addLast(v);

        for (int u: g.nodes.get(v).neighbours) {
            if (color[u] == 1) {
                System.out.println(q.size());
                for (int x: q) {
                    System.out.println(x + 1);
                }
                System.exit(0);
            } else if (color[u] == 0) {
                dfs(u);
            }
        }

        q.removeLast()  ;
        color[v] = 2;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        int n = in.nextInt();
        int m = in.nextInt();
        g = new Graph(n);
        for (int i = 0; i < m; ++i) {
            int u = in.nextInt() - 1, v = in.nextInt() - 1;
            g.nodes.get(u).neighbours.add(v);
        }

        color = new int[n];
        for (int i = 0; i < n; ++i) {
            if (color[i] == 0) {
                dfs(i);
            }
        }
    }

    static class Graph {
        ArrayList<Node> nodes = new ArrayList<>();

        Graph(int n) {
            for (int i = 0; i < n; ++i) {
                nodes.add(new Node());
            }
        }
    }

    static class Node {
        ArrayList<Integer> neighbours = new ArrayList<>();
    }
}