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.Scanner;
import java.util.ArrayDeque;
import java.util.ArrayList;
public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		int m=sc.nextInt();
		ArrayList<Integer>[] list=new ArrayList[n];
		for(int i=0; i<n; i++){
			list[i]=new ArrayList<>();
		}
		for(int i=0; i<m; i++){
			int a=sc.nextInt()-1;
			int b=sc.nextInt()-1;
			list[a].add(b);
		}
		int mini=Integer.MAX_VALUE;
		int index=-1;
		int[] ans=new int[n];
		for(int i=0; i<n; i++){
			ArrayDeque<Integer>  que=new ArrayDeque<>();
			int[] prev=new int[n];
			int[] dist=new int[n];
			boolean[] visited=new boolean[n];
			que.add(0);
			que.add(i);
			while(!que.isEmpty()){
				int t=que.poll();
				int j=que.poll();
				dist[j]=dist[t]+1;
				prev[j]=t;
				if(visited[j]){
					if(j==i && mini>dist[j]-1){
						index=i;
						mini=dist[j]-1;
						ans=prev;
					}
					break;
				}
				visited[j]=true;
				for(int k:list[j]){
					que.add(j);
					que.add(k);
				}
			}
		}
		if(mini!=Integer.MAX_VALUE){
			System.out.println(mini);
			int i=ans[index];
			System.out.println(index+1);
			while(i!=index){
				System.out.println(i+1);
				i=ans[i];
			}
		}else{
			System.out.println(-1);
		}
	}
}