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++ (GCC 9.2.1) (judge verdict: Accepted)

#ifdef __LOCAL
  #define _GLIBCXX_DEBUG
#endif
#include <bits/stdc++.h>
using namespace std;
template<typename T> bool chmax(T &a,T b) {if(a<b) {a=b; return true;} return false;}
template<typename T> bool chmin(T &a,T b) {if(a>b) {a=b; return true;} return false;}
#define itn int
#define fi first
#define se second
#define intmax numeric_limits<int>::max()
#define llmax numeric_limits<ll>::max()
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define rrep1(i,n) for(int i=(int)(n);i>=1;i--)
#define all(vec) vec.begin(),vec.end()
#define sortt(vec) sort((vec).begin(),(vec).end())
#define rsort(vec) sort((vec).rbegin(), (vec).rend())
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
typedef tuple<ll,ll,ll> tlll;
typedef tuple<int,int,int> tiii;
const ll mod=1e9+7;
const int inf=1<<30;
const ll lnf=1ll<<60;

int prevv[100010]; //頂点の最大数
struct edge{int to,cost;};

vector<ll> dijkstra(int n,vector<vector<edge> > G,int sv){
  fill(prevv,prevv+n,-1);
  priority_queue<pll,vector<pll>,greater<pll> > que;
  vector<ll> dist(n,lnf);
  dist[sv]=0;
  que.push(pll(0,sv));
  while(!que.empty()){
    ll cost; int v;
    tie(cost,v)=que.top();
    que.pop();
    if(dist[v]<cost) continue;
    for(int i=0;i<G[v].size();i++){
      edge e=G[v][i];
      ll ec=e.cost;
      int nv=e.to;
      if(dist[nv]>dist[v]+ec){
        dist[nv]=dist[v]+ec;
        prevv[nv]=v;
        que.push(pll(dist[nv],nv));
      }
    }
  }
  return dist;
}

//頂点への最短路
vector<int> get_path(int t){
  vector<int> path(0);
  for(;t!=-1;t=prevv[t]) path.push_back(t);
  reverse(path.begin(),path.end());
  return path;
}

int main(){
  int n,m; cin >> n >> m;
  vector<vector<edge> > to(n);
  vector<vector<int> > rto(n); 
  rep(i,m){
    itn a,b;
    cin >> a >> b;
    a--; b--;
    to[a].push_back({b,1});
    rto[b].push_back(a);
  }
  ll infdist=inf;
  itn thegv=-1;
  itn thesv=-1;
  rep(i,n){
    vector<ll> dist=dijkstra(n,to,i);
    for(auto v:rto[i]){
      if(chmin(infdist,dist[v]+1)){
        thegv=v;
        thesv=i;
      }
    }
  }
  if(thegv!=-1){
    vector<ll> dist=dijkstra(n,to,thesv);
    vector<int> path=get_path(thegv);
    cout << path.size() << endl;
    for(auto v:path){
      cout << v+1 << endl;
    }
    return 0;
  }
  cout << -1 << endl;
  return 0;
}