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>

using namespace std;
using llong = long long;
using ldbl = long double;
using lpair = pair<int, int>;

#define ALL(x) x.begin(), x.end()

const llong inf = 1'000'000'000'000'007ll;
const llong mod = 1e9+7;

int N, M;
vector<set<int>> edges;
vector<lpair> all_edge;
int target;

bool Heiro(vector<int> &ans) {
	static vector<int> visited(N+1, 0);

	if (ans.back() == target && visited[target] == 1) return true;

	vector<int> new_v;
	bool finded = false;
	for (auto edge : edges[ans.back()]) {
		if (visited[edge] == 1) return false;
		if (edge == target) finded = true;
		new_v.push_back(edge);
	}

	if (finded) {
		ans.push_back(-1);
		return true;
	}

	for (auto next : new_v) {
		ans.push_back(next);
		visited[next] = 1;
		if (Heiro(ans)) return true;
		ans.pop_back();
		visited[next] = 0;
	}
	return false;
}

set<int> BellmanFord() {
	vector<int> dis(N+1, 99999);
	dis[1] = 0;
	set<int> ret;
	for (int i = 0; i < 2*N; ++i) {
		for (auto edge : all_edge) {
			if (dis[edge.second] > dis[edge.first] - 1) {
				dis[edge.second] = dis[edge.first] - 1;
				if (i >= N) {
					edges[edge.first].insert(edge.second);
					ret.insert(edge.first);
				}
			}
		}
	}

	return ret;
}

int main() {
	cin >> N >> M;
	edges.resize(N+1);
	all_edge.resize(M);
	for (int i = 0; i < M; ++i) {
		int from, to;
		cin >> from >> to;
		all_edge[i] = lpair(from, to);
	}

	set<int> t_list = BellmanFord();

	if (t_list.size() == 0) {
		cout << -1 << endl;
		return 0;
	} else {
		for (auto t : t_list) {
			target = t;
			vector<int> ans;
			ans.push_back(target);
			Heiro(ans);
			ans.pop_back();
			if (ans.size() == 0) continue;
			cout << ans.size() << endl;
			for (auto v : ans)
				cout << v << endl;
			return 0;
		}
	}
	cout << -1 << endl;
	return 0;
}