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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define ll          long long
#define ld          long double

void dfs(ll x, ll p, vector<ll>&par, vector<ll>&vis, vector<vector<ll>>&g, vector<vector<bool>>&ed)
{
	vis[x]=1;
	par[x]=p;
	for(ll y : g[x])
	{
		if(vis[y]==1)
		{
			set<ll>cyc;
			ll cur=x;
			while(cur!=y)
			{
				cyc.insert(cur);
				cur=par[cur];
			}
			cyc.insert(cur);
			bool ok=1;
			// only el of set
			ll n=g.size()-1;
			for(ll z : cyc)
			{
				ll cnt=0;
				for(ll i=1; i<=n; i++) if(cyc.count(i)) cnt+=ed[z][i];
				if(cnt>1) {ok=0; break;}
			}
			if(ok)
			{
				cout<<cyc.size()<<endl;
				for(ll z : cyc) cout<<z<<endl;
				exit(0);
			}
		}
		else 
		{
			dfs(y, x, par, vis, g, ed);
		}
	}
	vis[x]=2;
}

void solve()  
{
	ll n, m; cin>>n>>m;
	vector<vector<ll>>g(n + 1);
	vector<vector<bool>>ed(n+1, vector<bool>(n+1, 0));
	for(ll i=0; i<m; i++) { ll x, y; cin>>x>>y; g[x].push_back(y); ed[x][y]=1; }
	vector<ll>par(n+1);
	vector<ll>vis(n+1, 0);
	for(ll i=1; i<=n; i++) if(vis[i]==0) { vis[i]=1; dfs(i, i, par, vis, g, ed); }
	cout<<-1<<endl;
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);  
	ll T = 1;
	//cin>>T; 
	while(T--)
	    solve();
	return 0;
}