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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <algorithm>
#include <set>
#include <map>
#include <bitset>
#include <cmath>
#define vv(a, b, c, d) vector<vector<d> >(a, vector<d>(b, c))
#define vvi std::vector<std::vector<int> >
#define vvl std::vector<std::vector<ll> >
#define MODs 1000000007;
typedef long long int ll;
typedef long double ld;
using namespace std;

//-------------------------------------------------------------------
ll MAX_V = 5000;

vvi G = vv(MAX_V, 0, 0, int);
vvi rG = vv(MAX_V, 0, 0, int);
vector<bool> used(MAX_V);
vector<ll> cmp(MAX_V);
vector<ll> vs;

void add_edge(ll from, ll to){ G[from].push_back(to), rG[to].push_back(from);}
void dfs(ll v){
  used[v] = true;
  for(int i=0;i<G[v].size();i++) if(!used[G[v][i]]) dfs(G[v][i]);
  vs.push_back(v);
}
void rdfs(ll v, ll k){
  used[v] = true;
  cmp[v] = k;
  for(int i=0;i<rG[v].size();i++) if(!used[rG[v][i]]) rdfs(rG[v][i], k);
}
int scc(ll V){
  for(int i=0;i<MAX_V;i++) used[i] = false;
  vs.clear();
  for(int i=0;i<=V;i++) if(!used[i]) dfs(i);
  for(int i=0;i<MAX_V;i++) used[i] = false;
  int k = 0;
  for(int i=vs.size()-1;i>=0;i--) if(!used[vs[i]]) rdfs(vs[i], k++);
  return k;
}

//-------------------------------------------------------------------
vector<ll> use(1000000, 0);
map<ll, ll> mp;
vector<ll> ans;
vector<int> p(2000, 0);

void df(ll from, ll now, ll c, ll d){
  mp.emplace(now, d);
  p[now] = from;
  set<ll> b;
  for(int i=0;i<G[now].size();i++) if(mp.find(G[now][i])!=mp.end()) b.insert(G[now][i]);
  if(b.size()==0){
    for(int i=0;i<G[now].size();i++){
      if(c!=cmp[G[now][i]]) continue;
      df(now, G[now][i], c, d+1);
    }
  }else{
    ll tmp = now;
    ans.push_back(tmp);
    while(b.find(tmp)==b.end()){
      tmp = p[tmp];
      ans.push_back(tmp);
    }
    std::cout << ans.size() << '\n';
    for(int i=0;i<ans.size();i++) std::cout << ans[i] << '\n';
    exit(0);
  }
  mp.erase(now);
}
int main(int argc, char const *argv[]) {
  ll N, M, a, b;
  std::cin >> N >> M;
  for(int i=0;i<M;i++){
    std::cin >> a >> b;
    add_edge(a, b);
  }
  scc(N+2);
  for(int i=1;i<=N;i++){
    df(-1, i, cmp[i], 1);
  }
  std::cout << -1 << '\n';
  return 0;
}