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 {

    static final int NONE = -1;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();

        List<List<Integer>> out = new ArrayList<>();
        boolean[][] in = new boolean[N][N];
        for (int i = 0; i < N; i++) out.add(new ArrayList<>());

        for (int i = 0; i < M; i++) {
            int u = sc.nextInt() - 1;
            int v = sc.nextInt() - 1;
            out.get(u).add(v);
            in[v][u] = true;
        }

        List<Integer> sol = null;

        for (int u = 0; u < N; u++) {
            List<Integer> cycle = shortestCycle(u, out, in[u]);
            if (cycle != null && (sol == null || cycle.size() < sol.size())) sol = cycle;
        }

        if (sol == null) System.out.println(-1);
        else {
            System.out.println(sol.size());
            for (int u : sol) System.out.println(u + 1);
        }
    }


    static List<Integer> shortestCycle(int s, List<List<Integer>> adj, boolean[] in) {
        int[] pred = new int[in.length];
        Arrays.fill(pred, NONE);
        Deque<Integer> deque = new ArrayDeque<>();
        deque.add(s);

        while (!deque.isEmpty()) {
            int u = deque.pollLast();

            for (int v : adj.get(u)) {
                if (pred[v] != NONE) continue;
                if (in[v]) {
                    List<Integer> cycle = new ArrayList<>();
                    while (pred[u] != NONE) {
                        cycle.add(u);
                        u = pred[u];
                    }
                    cycle.add(s);
                    cycle.add(v);
                    return cycle;
                }
                pred[v] = u;
                deque.addFirst(v);
            }
        }

        return null;
    }
}