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)

// 嘘解法です！！！！！

#include <cassert>
#include <cstdio>
#include <cmath>
#include <iostream>
#include <iomanip>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <numeric>
#include <algorithm>

using namespace std;
using lint = long long;
constexpr int MOD = 1000000007, INF = 1010101010;
constexpr lint LINF = 1LL << 60;

template <class T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
	for (const auto &e : vec) os << e << (&e == &vec.back() ? "\n" : " ");
	return os;
}

#ifdef _DEBUG
template <class T>
void dump(const char* str, T &&h) { cerr << str << " = " << h << "\n"; };
template <class Head, class... Tail>
void dump(const char* str, Head &&h, Tail &&... t) {
	while (*str != ',') cerr << *str++; cerr << " = " << h << "\n";
	dump(str + (*(str + 1) == ' ' ? 2 : 1), t...);
}
#define DMP(...) dump(#__VA_ARGS__, __VA_ARGS__)
#else 
#define DMP(...) ((void)0)
#endif

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

int main() {

	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	int N, M;
	cin >> N >> M;

	vector<vector<int>> d(N, vector<int>(N, INF));
	for (int i = 0; i < M; i++) {
		int a, b;
		cin >> a >> b;
		a--, b--;
		d[a][b] = 1;
	}
	DMP(d);

	for (int k = 0; k < N; k++) {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				if (d[i][k] == INF || d[k][j] == INF) continue;
				d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
			}
		}
	}
	DMP(d);

	int mincyc = INF;
	vector<int> ans;
	for (int i = 0; i < N; i++) chmin(mincyc, d[i][i]);
	if (mincyc == INF) cout << -1 << "\n";
	else {

		int st = -1;
		for (int i = 0; i < N; i++) {
			if (d[i][i] == mincyc) {
				st = i;
				break;
			}
		}

		while (ans.size() < mincyc) {

			int now = st;
			for (int i = 0; i < N; i++) {
				if (d[i][i] == mincyc && d[now][i] == 1 && mincyc == d[i][st] + d[st][i]) {
					ans.emplace_back(i);
					now = i;
					break;
				}
			}

		}

		cout << ans.size() << "\n";
		for (const auto &e : ans) cout << e + 1 << "\n";
	}

	return 0;
}