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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file 
#include <ext/pb_ds/tree_policy.hpp> 
#include <functional> // for less 
#define pb push_back
#define For(i,s,e) for (ll i=(s); i<(e); i++)
#define Debug_array(a,n) for (ll i=(0); i<(n); i++) cout<<a[i]<<" "
#define Foe(i,s,e) for (ll i=(s); i<=(e); i++)
#define Fod(i,s,e) for (ll i=(s)-1; i>=(e); i--)
#define pii pair<ll,ll>
#define fi first
#define se second
#define endl "\n"
#define mp make_pair
#define big_prime 15486277
#define bigger_prime 179424697
#define biggest_prime 32416188691
//#define random_shuffle(indices.begin(), indices.end());
//std::random_device rd;
//std::mt19937 g(rd());
//std::shuffle(v.begin(), v.end(), g);
using namespace __gnu_pbds; 
using namespace std;
template <class T> ostream& operator << (ostream &os, const vector<T> &v) { for (T i : v) os << i << ' '; return os; }
template <class T> ostream& operator << (ostream &os, const set<T> &v) { for (T i : v) os << i << ' '; return os; }
template <class T, class S> ostream& operator << (ostream &os, const pair<T, S> &v) { os << v.first << ' ' << v.second; return os; }
// template <class T, class S> ostream& operator << (ostream &os, const unordered_map<T, S> &v) { for (auto i : v) os << '(' << i.first << "=>" << i.second << ')' << ' '; return os; }
 
#ifndef ONLINE_JUDGE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <class Arg1> void __f(const char* name, Arg1&& arg1) { cerr << name << " : " << arg1 << endl; }
    template <class Arg1, class... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args) {
        const char* sep = strchr(names + 1, ',');
        cerr.write(names, sep - names) << " : " << arg1 << "  ";
        __f(sep + 1, args...);
    }
#else
#define trace(...) 0
#define _CRT_SECURE_NO_WARNINGS
#endif // ifndef ONLINE_JUDGE


typedef long long ll;
typedef tree<ll, null_type, less<ll>, rb_tree_tag,tree_order_statistics_node_update> ordered_set; 

#define Mod 1000000007
#define MAX 1005

ll n,m;
vector<ll> adj[MAX];

bool ans;
vector<ll> final;
ll root;
ll visited[MAX];

bool f(ll s,bool ok){
	visited[s]=1;

	bool no_back = 1;
	for(ll x:adj[s])
	{
		if(x==root) continue;
		if(visited[x]==2) continue;
		if(visited[x]==1) no_back = 0;
	}

	for(ll x:adj[s])
	{
		if(visited[x]==1 && x==root && ok && no_back)
		{
			ans = 1;
			final.pb(s);
			return 1;
		} 
	}	

	for(ll x:adj[s]) if(visited[x]==0)
	{
		if(f(x,ok && no_back)){
			final.pb(s);
			return 1;
		}
	}

	visited[s]=2;
	return 0;
}


vector<ll> components[MAX];

int main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	memset(visited,0,sizeof(visited));
	ans = 0;
	cin>>n>>m;
	For(i,0,m){
		ll x,y;cin>>x>>y;x--;y--;
		adj[x].pb(y);
	}

	For(start,0,n)
	{
		memset(visited,0,sizeof(visited));
		root = start;
		f(root,1);
		if(ans){
			cout<<final.size()<<endl;
			for(ll x:final){
				cout<<x+1<<endl;
			}
			return 0;
		}
	}
	cout<<-1<<endl;
	return 0;





























	
}