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;
#define mp       make_pair
#define pb       push_back
#define all(x)   (x).begin(),(x).end()
#define YES() printf("YES\n")
#define NO() printf("NO\n")
#define Yes() printf("Yes\n")
#define No() printf("No\n")
#define in(x,y,h,w) x >= 0 && x < h && y >= 0 && y < w

#define int long long
//using ll = long long;
using P = pair<int,int>;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
 
const int INF=1e+9;
const double EPS=1e-9;
const int MOD=1000000007;

const int dx[]={1,0,-1,0},dy[]={0,-1,0,1};

vector<vector<int>> G;
int root[1010],depth[1010],par[1010];

void dfs(int v,int d,int r,int p){
	depth[v] = d;
	root[v] = r;
	par[v] = p;
	for(int to : G[v]){
		if(root[to] == -1) dfs(to,d + 1,r,v);
	}
}

signed main(){
	int n,m;
	cin >> n >> m;
	G.resize(n);
	for(int i = 0;i < m;i++){
		int a,b;
		cin >> a >> b; a--;b--;
		G[a].push_back(b);
	}
	for(int i = 0;i < n;i++) root[i] = -1;
	for(int i = 0;i < n;i++){
		if(root[i] == -1) dfs(i,0,i,-1);
	}
	for(int i = 0;i < n;i++){
		for(int to : G[i]){
			if(!(root[i] == root[to] && depth[i] > depth[to])) continue;
			bool isCycle = true;
			int pos = i;
			vector<int> ans;
			while(pos != to){
				ans.push_back(pos);
				for(int to2 : G[pos]){
					if(root[i] == root[to2] && !(pos == i && to2 == to) && depth[to2] < depth[pos] && depth[to] <= depth[to2]) isCycle = false;
				}
				pos = par[pos];
			}
			if(isCycle){
				ans.push_back(to);
				cout << ans.size() << endl;
				for(int v : ans) cout << v + 1 << endl;
				return 0;
			}
		}
	}
	cout << -1 << endl;
}