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 <iostream>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <set>
#include <string>
#include <map>
#include <list>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <utility>
using namespace std;

typedef long long int64;
typedef unsigned long long uint64;
typedef pair<int, int> ii;
typedef pair<int64, int64> int64int64;
typedef vector<int> vint;
typedef vector<int64> vint64;
typedef vector<uint64> vuint64;
typedef vector<vint> vvint;
typedef vector<vint64> vvint64;
typedef vector<vuint64> vvuint64;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;

#define mp make_pair
#define pb push_back
#define ALL(a) a.begin(), a.end()
#define fi first
#define se second
#define fori(n) for(int i = 0; i < n; ++i)
#define forj(n) for(int j = 0; j < n; ++j)
#define fork(n) for(int k = 0; k < n; ++k)

template<typename T>
void prinMatrix(string txt, vector<vector<T>> &a){
	cout << txt;
	fori(a.size()){
		forj(a[i].size())
			cout << a[i][j] << ' ';

		cout << endl;
	}
}

template<typename T>
void printVector(string txt, vector<T> &a){
	cout << txt;
	for (int i = 0; i < a.size(); ++i)
		cout << a[i] << ' ';
	cout << endl;
}

int n, m;
vvii g;
vint vis;
vint dist;
vint st;
vint ans;

#define INF (int)(1e9)

bool met = false;

int dfs(int v, int len){
	if (vis[v] == 1){
		ans.pb(v);
		return 2;
	}
	vis[v] = 1;
	dist[v] = len;

	st.push_back(v);
	fori(g[v].size())
		g[v][i].second = dist[g[v][i].first];

	sort(ALL(g[v]), [](ii &a, ii &b){return a.second > b.second; });

	fori(g[v].size()){
		int to = g[v][i].first;
		if (vis[to] == 1){
			int k = dfs(to, len + 1);
			if (k == 2){
				if (v == ans[0])
					met = true;

				if (!met)
					ans.pb(v);
				st.pop_back();
				return 2;
			}
		}
	}

	fori(g[v].size()){
		int to = g[v][i].first;
		if (vis[to] == 0){
			int k = dfs(to, len + 1);
			if (k == 2){
				if (v == ans[0])
					met = true;

				if (!met)
					ans.pb(v);
				st.pop_back();
				return 2;
			}
		}
	}

	vis[v] = 2;
	st.pop_back();
	return 0;
}

int main(){

	ios_base::sync_with_stdio(0);
	cin.tie();

	cin >> n >> m;
	
	g.resize(n);
	vis.resize(n, 0);
	dist.resize(n, 0);
	fori(m){
		int a, b;
		cin >> a >> b;
		--a; --b;
		g[a].pb({ b, INF });
	}

	if (m == 0){
		cout << -1 << endl;
		return 0;
	}

	fori(n){
		if (!vis[i]){
			met = false;
			
			int k = dfs(i, 1);
			if (k == 2){
				cout << ans.size() << endl;
				reverse(ALL(ans));
				forj(ans.size()){
					cout << ans[j] + 1 << endl;
				}
				return 0;
			}

			fill(ALL(vis), 0);
		}
	}

	cout << -1 << endl;

	return 0;
}