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.List;
import java.util.Scanner;
 
public class Main {
    static Scanner sc;
    static List<List<Integer>> t;
    static boolean[] visited;
    static boolean[] grey;
    static int[] prev;
    static List<Integer> shortestCycle;
    public static void main(String[] args) {
        sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        t = new ArrayList<List<Integer>>(N);
        for(int i = 0; i <= N; i++) {
            t.add(new ArrayList<Integer>());
        }
 
        for(int i = 0; i < M; i++) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            t.get(a).add(b);
        }
        visited = new boolean[N+1];
        grey = new boolean[N+1];
        prev = new int[N+1];
 
        for(int i = 1; i <= N; i++) {
            dfs(i);
 
            visited = new boolean[N+1];
            grey = new boolean[N+1];
            prev = new int[N+1];
        }
        if(shortestCycle != null) {
            System.out.println(shortestCycle.size());
            for(int l: shortestCycle) {
                System.out.println(l);
            }
        } else {
            System.out.println("-1");
        }
    }
 
    static void dfs(int s) {
        grey[s] = true;
        for(int i : t.get(s)) {
            if(!visited[i]) {
                if(grey[i]) {
                    List<Integer> lst = new ArrayList<>();
                    int tmp = s;
                    while(tmp != i) {
                        lst.add(tmp);
                        tmp = prev[tmp];
                    }
                    lst.add(i);
                    if(shortestCycle == null || lst.size() < shortestCycle.size()) {
                        shortestCycle = lst;
                    }
                } else {
                    prev[i] = s;
                    dfs(i);
                }
            }
        }
        visited[s] = true;
    }
}