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.Scanner;
import java.util.Arrays;
import java.lang.Math;
import java.util.ArrayList;
import java.util.List;
import java.util.Stack;
  
public class Main {
	static class Graph {
      	Stack mStack = new Stack();
        int mVertices;
      	ArrayList<ArrayList<Integer>> mEdges = new ArrayList<ArrayList<Integer>>();
      
      	Graph(int n) {
          mVertices = n;
          for (int i = 0; i < n; i++) {
          	mEdges.add(new ArrayList<Integer>());
          }
        }
      
      	void add(int a, int b) {
          mEdges.get(a - 1).add(b - 1); 
        }
      
      	boolean iSCyclic() {
          boolean[] visited = new boolean[mVertices];
          //do DFS from each node
          for (int i = 0; i < mVertices; i++) {
          	if (isCyclicHelper(i, visited, mStack))
              return true;
          }
          return false;
        }
      
      	boolean isCyclicHelper(int a, boolean[] visited, Stack stack) {
          visited[a] = true;
          stack.push(a);
          
          for (Integer b: mEdges.get(a)) {
            if (!visited[b] && isCyclicHelper(b, visited, stack))
              return true;
            else if(stack.contains(b))
              return true;
          }
          
          stack.pop();
          return false;
        }
      
      	void printResult() {
          //Object[] vals = mStack.toArray();
          System.out.println(mStack.size());
          while (!mStack.isEmpty()) {
              System.out.println((Integer)mStack.pop() + 1);
          }
          //System.out.println(Arrays.toString(mStack.toArray()));
        }
      }

    public static void main(String[] args) {
       
      Scanner scanner = new Scanner(System.in);
      int n = scanner.nextInt();
      int m = scanner.nextInt();
      
      Graph g = new Graph(n);
      for (int i = 0; i < m; i++) {
      	int a = scanner.nextInt();
      	int b = scanner.nextInt();
        g.add(a, b);
      }
      g.iSCyclic();
      g.printResult();
    }
  
}