Problem:
Score : 600 points

Problem Statement

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.

Notes

For a directed graph G = (V, E), we call a directed graph G' = (V', E') satisfying the following conditions an induced subgraph of G:

V' is a (non-empty) subset of V.

E' is the set of all the edges in E that have both endpoints in V'.

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.

Sample Input 1

4 5
1 2
2 3
2 4
4 1
4 3

Sample Output 1

3
1
2
4

The induced subgraph of G whose vertex set is \{1, 2, 4\} has the edge set \{(1, 2), (2, 4), (4, 1)\}. The in-degree and out-degree of every vertex in this graph are both 1.

Sample Input 2

4 5
1 2
2 3
2 4
1 4
4 3

Sample Output 2

-1

There is no induced subgraph of G that satisfies the condition.

Sample Input 3

6 9
1 2
2 3
3 4
4 5
5 6
5 1
5 2
6 1
6 2

Sample Output 3

4
2
3
4
5

Solution in Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var sc, wr = bufio.NewScanner(os.Stdin), bufio.NewWriter(os.Stdout)

func scanString() string   { sc.Scan(); return sc.Text() }
func scanRunes() []rune    { return []rune(scanString()) }
func scanInt() int         { a, _ := strconv.Atoi(scanString()); return a }
func scanInt64() int64     { a, _ := strconv.ParseInt(scanString(), 10, 64); return a }
func scanFloat64() float64 { a, _ := strconv.ParseFloat(scanString(), 64); return a }

func scanInts(n int) []int {
	res := make([]int, n)
	for i := 0; i < n; i++ {
		res[i] = scanInt()
	}
	return res
}

func debug(a ...interface{}) { fmt.Fprintln(os.Stderr, a...) }

func abs(a int) int {
	if a < 0 {
		return -a
	}
	return a
}
func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}
func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

//•*¨*•.¸¸♪main•*¨*•.¸¸♪(　-ω-)ノ　(　・ω・)
func main() {
	defer wr.Flush()
	sc.Split(bufio.ScanWords)
	sc.Buffer(make([]byte, 10000), 1001001)

	n := scanInt()
	m := scanInt()
	g := make([][]int, n)
	for i := 0; i < m; i++ {
		a, b := scanInt()-1, scanInt()-1
		g[a] = append(g[a], b)
	}

	var dfs func(a int, arr []int) bool
	done := make([]bool, n)
	var ans []int
	dfs = func(a int, arr []int) bool {
		if len(arr) != 1 && a == arr[0] {
			ans = arr
		}
		for _, b := range g[a] {
			if done[b] {
				continue
			}
			done[b] = true
			if dfs(b, append(arr, b)) {
				return true
			}
			done[b] = false
		}
		return false
	}

	for i := 0; i < n; i++ {
		if dfs(i, []int{i}) {
			break
		}
	}

	if len(ans) == 0 {
		fmt.Fprintln(wr, -1)
	} else {
		for {
			f := make([]bool, n)
			for i := 0; i < len(ans); i++ {
				f[ans[i]] = true
			}
			a, b := -1, -1
			for i := 0; i < len(ans)-1; i++ {
				for _, v := range g[ans[i]] {
					if f[v] && v != ans[i+1] {
						a = i
						for j := 0; j < len(ans)-1; j++ {
							if v == ans[j] {
								b = j
								break
							}
						}
					}
				}
				if b != -1 {
					break
				}
			}
			debug(ans, a, b)
			if b == -1 {
				fmt.Fprintln(wr, len(ans)-1)
				for i := 0; i < len(ans)-1; i++ {
					fmt.Fprintln(wr, ans[i]+1)
				}
				return
			}
			if a > b {
				ans = append(ans[b:a+1], ans[b])
			} else {
				ans = append(ans[:a+1], ans[b:]...)
			}
		}
	}
}