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<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=1005;
int N,M,vis[MAXN],F[MAXN];
int Cnt[MAXN][MAXN];
int cnt[MAXN][MAXN];
int nt[MAXN][MAXN];
vector<int>P[MAXN];
queue<int>Q;
int Ans[MAXN],Len;
void Print(int x){
	if(vis[x]){
		printf("%d\n",Len);
		for(int i=1;i<=Len;i++)
			printf("%d\n",Ans[i]);
		return ;
	}
	vis[x]=1;Ans[++Len]=x;Print(F[x]);
}
int BFS(int u,int Pt){
	while(!Q.empty())Q.pop();
	Q.push(u);vis[u]=1;
	while(!Q.empty()){
		u=Q.front();Q.pop();
		int size=P[u].size();
		int Ok=0,flag=0;
		for(int i=0;i<size;i++){
			int v=P[u][i];
			if(v==Pt&&!Ok)Ok=1;
			else if(vis[v]){flag=1;continue;}
			vis[v]=1;F[v]=u;
			if(nt[u][v])continue;
			Q.push(v);
		}
		if(Ok&&!flag){
			F[Pt]=u;
			memset(vis,0,sizeof(vis));
			Print(Pt);
			return 1;
		}
	}
	return 0;
}
int main(){
	scanf("%d%d",&N,&M);
	for(int i=1,x,y;i<=M;i++){
		scanf("%d%d",&x,&y);
		if(x==y)continue;
		P[x].push_back(y);
		cnt[x][y]++;Cnt[x][y]++;
		cnt[y][x]++;
	}
	for(int i=1;i<=N;i++)
		for(int j=i+1;j<=N;j++){
			if(cnt[i][j]==2&&Cnt[i][j]==1){
				printf("2\n%d\n%d\n",i,j);
				return 0;
			}
			if(cnt[i][j]>=2)
				nt[i][j]=1;
		}
	for(int i=1;i<=N;i++){
		memset(vis,0,sizeof(vis));
		if(BFS(i,i))return 0;
	}
	printf("-1\n");
}