Problem: 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.

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.

Examples

Input

4 5
1 2
2 3
2 4
4 1
4 3


Output

3
1
2
4


Input

4 5
1 2
2 3
2 4
1 4
4 3


Output

-1


Input

6 9
1 2
2 3
3 4
4 5
5 6
5 1
5 2
6 1
6 2


Output

4
2
3
4
5

Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Scanner;

public class Main {
	static class Vertex{
		List<Integer> adj;
		public Vertex() {
			adj = new ArrayList<Integer>();
		}
	}
	static List<Integer> findCircuit(Vertex vs[] , int startPoint){
		Vertex v = vs[startPoint];
		if(v.adj.isEmpty()){
			return null;
		}
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(startPoint);
		int N = vs.length;
		int prev[] = new int[N];
		Arrays.fill(prev, -1);
		while(!q.isEmpty()){
			int cur = q.poll();
			Vertex vcur = vs[cur];
			for(int n : vcur.adj){
				if(n == startPoint){ // find circuit
					List<Integer> ret = new ArrayList<Integer>();
					while(prev[cur] != startPoint){
						ret.add(cur);
						cur = prev[cur];
					}
					ret.add(cur);
					ret.add(startPoint);					
					return ret;
				}
				if(prev[n] == -1){
					prev[n] = cur;
					q.add(n);
				}
			}
		}
		return null;
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		Vertex vs[] = new Vertex[N];
		for(int i = 0 ; i < vs.length ; ++i){
			vs[i] = new Vertex();
		}
		for(int i = 0 ; i < M ; ++i){
			int a = sc.nextInt() - 1;
			int b = sc.nextInt() - 1;
			vs[a].adj.add(b);
		}
		List<Integer> minCircuit = null;
		for(int i = 0 ; i < N ; ++i){
			List<Integer> circuit = findCircuit(vs, i);
			if(circuit != null){
				if(minCircuit == null){
					minCircuit = circuit;
				}else if(circuit.size() < minCircuit.size()){
					minCircuit = circuit;
				}
			}
		}
		if(minCircuit == null){
			System.out.println(-1);			
		}else{
			System.out.println(minCircuit.size());
			for(int v : minCircuit){
				System.out.println(v + 1);
			}
		}
	}
}