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

//=====I/O=====

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

func scanInt() int {
	sc.Scan()
	a,_ := strconv.Atoi(sc.Text())
	return a
}

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

func scanText() string {
	sc.Scan()
	return sc.Text()
}

func printInts(a ...int) {
	for i, e := range a {
		fmt.Fprint(wr, e)
		if i != len(a)-1 { fmt.Fprint(wr, " ") }
	}
	fmt.Fprintln(wr)
	wr.Flush()
}

//=====main=====

func main() {
	sc.Split(bufio.ScanWords)

	n, m := scanInt(), scanInt()
	g := make([][]int, n)

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

	r := make([]int, 0)
	used := make([]bool, n)
	if ok, arr, v := next(0, r,used,g); ok {
		// fmt.Println(arr,v)
		f := 0
		for i := 0; i < len(arr); i++ {
			if arr[i] == v {
				f = i
			}
		}

		fmt.Println(len(arr)-f)

		for i := f; i < len(arr); i++ {
			fmt.Println(arr[i]+1)
		}
	} else {
		fmt.Println(-1)
	}

}

func next(a int, r []int, used []bool, g [][]int) (bool, []int, int) {
	used[a] = true
	r = append(r, a)
	for _, e := range g[a] {
		if used[e] {
			return true, r, e
		}

		if ok, arr, v := next(e, r,used,g); ok {
			return ok, arr, v
		}
	}
	used[a] = false
	r = r[:len(r)-1]

	return false, make([]int, 0), -1
}