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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Queue;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		int m = Integer.parseInt(sa[1]);
		List<List<Integer>> list = new ArrayList<>(n);
		for (int i = 0; i < n; i++) {
			list.add(new ArrayList<>());
		}
		for (int i = 0; i < m; i++) {
			sa = br.readLine().split(" ");
			int a = Integer.parseInt(sa[0]) - 1;
			int b = Integer.parseInt(sa[1]) - 1;
			list.get(a).add(b);
		}
		br.close();

		int ans = n + 1;
		int idx = -1;
		for (int i = 0; i < n; i++) {
			Queue<Integer> que = new ArrayDeque<>();
			que.add(i);
			int[] d = new int[n];
			Arrays.fill(d, n + 1);
			d[i] = 0;
			label:
			while (!que.isEmpty()) {
				Integer cur = que.poll();
				List<Integer> list2 = list.get(cur);
				for (int next : list2) {
					int alt = d[cur] + 1;
					if (next == i) {
						if (alt < ans) {
							ans = alt;
							idx = i;
						}
						break label;
					}
					if (d[next] == n + 1) {
						que.add(next);
						d[next] = alt;
					}
				}
			}
		}
		if (ans == n + 1) {
			System.out.println(-1);
		} else {
			System.out.println(ans);
			Queue<List<Integer>> que = new ArrayDeque<>();
			List<Integer> listA = new ArrayList<>();
			listA.add(idx);
			que.add(listA);
			while (!que.isEmpty()) {
				List<Integer> cur = que.poll();
				List<Integer> list2 = list.get(cur.get(cur.size() - 1));
				for (int next : list2) {
					if (next == idx) {
						for (int i : cur) {
							System.out.println(i + 1);
						}
						return;
					}
					List<Integer> listN = new ArrayList<>(cur.size() + 1);
					listN.addAll(cur);
					listN.add(next);
					que.add(listN);
				}
			}
		}
	}
}