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>
#define ld long double
#define endl "\n"
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pb(x) push_back(x)
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define rep(i,l,r) for(int i = (int)l;i<(int)r;i++)
#define td(v) v.begin(),v.end()
#define sz(v) (int)v.size()
//#define M   1000000007 // 1e9 + 7
//#define int long long
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
inline ll mod(ll n, ll m){ ll ret = n%m; if(ret < 0) ret += m; return ret; }
ll gcd(ll a, ll b){return (b == 0LL ? a : gcd(b, a%b));}
ll exp(ll a,ll b,ll m){
    
    if(b==0LL) return 1LL;
    if(b==1LL) return mod(a,m);
    ll k = mod(exp(a,b/2,m),m);
    if(b&1LL){
        return mod(a*mod(k*k,m),m);
    }
    else return mod(k*k,m);
}

const int N = 1010;
vector<int> g[N];
int pai[N];
int d[N];
int n,m;
int vis[N];
vector<int> ans;
void bfs(int ini){
  queue<int> q;
  q.push(ini);
  d[ini]=0;
  int small = 1e9;
  int id=0,id2=0;
  vis[ini] = 1;
  while(!q.empty()){
    int cur = q.front();
      q.pop();
      for(int to : g[cur]){
          if(!vis[to]){
            vis[to] = 1;
            q.push(to);
            d[to] = d[cur] + 1;
            pai[to] = cur;
          }
          if(to==ini){
            vector<int> res;
            int c = cur;
            while(c!=to){
              res.pb(c);
              c = pai[c];
            }
            res.pb(to);
            if(ans.size()==0 or res.size() < ans.size())ans = res;
            return;
          }
      }
  }
 
   
}

// long long ? ?
int32_t main(){
  fastio;
  cin>>n>>m;
  for(int i=0;i<m;i++){
    int a,b;
    cin>>a>>b;
    g[a].pb(b);
  }
  for(int i=1;i<=n;i++){
    for(int j=1;j<=n;j++){
      vis[j] = 0;
      d[j] = 0;
      pai[j]= -1;
    }
    bfs(i);
  }
  if(ans.size()){
    cout<<ans.size()<<endl;
    for(int x : ans){
      cout<<x<<endl;
    }
    return 0;
  }
  cout<<-1<<endl;



  // Math -> gcd it all
  // Did u check N=1? Did you switch N,M?
}