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 _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)

typedef long long ll;
typedef pair<ll, ll> P;
const ll INF = 1LL<<60;

ll gcd(ll a, ll b) {
  if(b == 0) return a;
  return gcd(b, a % b);
}

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

ll modpow(ll a, ll n, ll mod) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}

vector<bool> seen, finished;

int pos = -1;
stack<int> hist;

void dfs(const vector<vector<int>> &G, int v, int p){

  seen[v] = true;
  hist.push(v);

  for(auto nv : G[v]){
    if(nv == p) continue;

    if(finished[nv]) continue;

    if(seen[nv] && !finished[nv]){
      pos = nv;
      return;
    }

    dfs(G, nv, v);

    if(pos != -1) return;

  }

  hist.pop();
  finished[v] = true;


}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int N, M;
  cin >> N >> M;
  vector<vector<int>> G(N);

  int a, b;
  rep(i, M){
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
  }

  seen.assign(N, false), finished.assign(N, false);

  dfs(G, 0, -1);
  
  set<int> cycle;
  while(!hist.empty()){
    int t = hist.top();
    cycle.insert(t);
    hist.pop();
    if(t == pos) break;
  }

  if(cycle.size() == 0){
    cout << -1 << endl;
  }else{
    cout << cycle.size() << endl;
    for(auto itr = cycle.begin(); itr != cycle.end(); itr++){
      cout << (*itr) +1 << endl;
    }
  }


  return 0;
}