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)

//#pragma GCC optimize ("-O3","unroll-loops")
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>
#include<bitset>
#include<limits.h>
#include<cassert>
#include<iterator>
#include<complex>
#include<stack>


#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v,n) reverse(v,v+n);
#define VREVERSE(v) reverse(v.begin(), v.end());
#define ll long long
#define pb(a) push_back(a)
#define m0(x) memset(x,0,sizeof(x))
#define print(x) cout<<x<<'\n';
#define pe(x) cout<<x<<" ";
#define lb(v,n) lower_bound(v.begin(), v.end(), n);
#define ub(v,n) upper_bound(v.begin(), v.end(), n);
#define int long long
#define all(x) (x).begin(), (x).end()
//#define double long double

using namespace std;

template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

const int MOD = 1e9 + 7;
const ll INF = 1e17;
const int INT_INF = 1e9;
const double pi = acos(-1);
const double EPS = 1e-10;
typedef pair<int, int>P;
const int MAX = 200020;

vector<int>G[1010];
int N,M;
int ans;
int s;
void BFS(int x) {
	vector<bool>used(N+1);
	used[x] = true;
	queue<int>que;
	int cnt = 1;
	que.push(x);
	vector<int>nex;
	//cout << "v:" << x << endl;
	while (!que.empty()) {
		int now = que.front(); que.pop();
		//pe(now);
		used[now] = true;
		for (auto v : G[now]) {
			if (v == x) {
				pe(v); print(cnt);
				if (ans > cnt) {
					s = v;
					ans = cnt;
				}
				return;
			}
			if (used[v])continue;
			nex.pb(v);
			used[v] = true;
		}
		
		if (que.empty()) {
			REP(i, nex.size()) {
				que.push(nex[i]);
			}
			nex.clear();
			cnt++;
		}
	}
}
//bool done = false;
//bool used[1010];
//void DFS(int n,vector<int>vec) {
//	if (done)return;
//	used[n] = true;
//	for (auto nex : G[n]) {
//		if (nex == n) {
//			REP(i, vec.size()) {
//				print(vec[i]);
//			}
//			done = true;
//			return;
//		}
//		if (used[nex])continue;
//		vector<int>tmp = vec;
//		tmp.pb(nex);
//		DFS(nex, tmp);
//	}
//}
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	ans = 1e9;
	cin >> N >> M;
	REP(i, M) {
		int u, v; cin >> u >> v;
		//u--, v--;
		G[u].pb(v);
	}
	FOR(i,1, N+1) {
		BFS(i);
	}
	if (ans == 1e9) {
		print(-1);
	}
	else {
		print(ans);
		vector<int>pre(N + 1);
		queue<int>que;
		que.push(s);
		vector<int>used(N + 1);
		while (!que.empty()) {
			int now = que.front(); que.pop();
			used[now] = true;
			for (auto nex : G[now]) {
				if (pre[nex] == 0) {
					pre[nex] = now;
				}
				if (nex == s) {
					int pos = now;
					print(pos);
					while (true) {
						pos = pre[pos];
						print(pos);
						if (pos == s)return 0;
					}
				}
				if (used[nex])continue;
				que.push(nex);
			}
		}
	}
}