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.*;
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		g = new ArrayList[n];
		for(int i = 0; i < n; i++) {
			g[i] = new ArrayList<>();
		}
		for(int i = 0; i < m; i++) {
			int u = sc.nextInt()-1;
			int v = sc.nextInt()-1;
			g[u].add(v);
		}
		prev = new int[n][n];
		dist = new int[n][n];
		int ucycle = -1; int dmin = n+1;
		for(int i = 0; i < n; i++) {
			Arrays.fill(prev[i], -1);
			Arrays.fill(dist[i], -1);
			int d = bfs(i);
			if(d > 0 && d < dmin) {
				ucycle = i;
				dmin = d;
			}
		}
		if(ucycle == -1) {
			System.out.println(-1);
		}
		else {
			StringBuilder sb = new StringBuilder();
			sb.append(dmin+"\n");
			ArrayList<Integer> vs = new ArrayList<Integer>();
			int i = prev[ucycle][ucycle];
			while(i != ucycle) {
				vs.add(i);
				i = prev[ucycle][i];
			}
			vs.add(ucycle);
			Collections.reverse(vs);
			for(int v: vs) {
				sb.append((v+1)+"\n");
			}
			System.out.print(sb.toString());
		}
	}
	static ArrayList<Integer>[] g;
	static int[][] prev;
	static int[][] dist;
	static int bfs(int i) {
		LinkedList<Integer> q = new LinkedList<>();
		q.add(i);
		dist[i][i] = 0;
		while(!q.isEmpty()) {
			int u = q.poll();
			for(int v: g[u]) {
				if(v == i) {
					prev[i][i] = u;
					return dist[i][u]+1;
				}
				else if(dist[i][v] < 0) {
					dist[i][v] = dist[i][u]+1;
					prev[i][v] = u;
					q.add(v);
				}
			}
		}
		return -1;
	}
}