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("O4")
#include <bits/stdc++.h>
#define For(i,a,b) for(int i=a;i<=b;i++)
#define F first
#define S second
#define all(x) x.begin(),x.end()
#define sz(x) ((int)x.size())
#define MOD (ll)(1e9+7)
#define INF 1e17
#define int ll
#define EPS (1e-6)
using namespace std;
using ll=long long;
using pii=pair<int,int>;

int gcd(int a,int b) { return b==0?a:gcd(b,a%b); }
int lcm(int a,int b) { return a/gcd(a,b)*b; }

int fpow(int b,int p)
{
	if(p==0) return 1;
	int h=fpow(b,p/2);
	if(p%2) return h*b%MOD*h%MOD;
	return h*h%MOD;
}

struct Node{
	vector<int> con;
	int tag;
	Node(){
		tag=-1;
		con.clear();
	}
}grp[1050];
int nt=1;
set<int> st;
int dfs(int now){
	st.insert(now);
	grp[now].tag=nt; nt++;
	
	int mt=0,id=0;
	for(auto &i:grp[now].con){
		if(grp[i].tag>mt){
			mt=grp[i].tag;
			id=i;
		}
	}
	if(id!=0) return id;
	
	for(auto &i:grp[now].con){
		int t=dfs(i);
		if(t!=-1) return t;
	}
	
	grp[now].tag=-1; nt--;
	st.erase(now);
	return -1;
}

int32_t main() {
	//ios::sync_with_stdio(false);
	//cin.tie(0);
	
	int n,m;; cin>>n>>m;
	int s,t;
	For(i,0,m-1){
		cin>>s>>t;
		grp[s].con.emplace_back(t);
	}
	grp[1].tag=0;
	int ans=dfs(1);
	if(ans==-1){
		cout<<"-1\n";
		return 0;
	}
	int now=ans;
	vector<int> out;
	while(sz(out)==0 || now!=ans){
		//cout<<now<<"\n";
		//int mt=grp[grp[now].con[0]].tag,id=grp[now].con[0];
		int mt=0,id=0;
		for(auto &i:grp[now].con){
			if(!st.count(i)) continue;
			if(grp[i].tag>mt){
				id=i;
				mt=grp[i].tag;
			}
		}
		out.emplace_back(id);
		now=id;
	}
	cout<<sz(out)<<"\n";
	for(auto &i:out) cout<<i<<"\n";
	return 0;
}