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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;
inline void prep ()
{
  cin.tie (0);
  cin.sync_with_stdio (0);
};
const double pi = acos(-1.);
//printf(%.12lf\n,mid);
// __builtin_popcount(int) count 1's in binary
// get decimal part of number
/*
double intpart;
double thiss = modf((double)(n*m)/(double)k, &intpart);
*/
// round = (int)(num+.5)
int mod = 998244353;

long long mymod(long long num){
  if (num< 0 && !(num%mod == 0)){
    if (mod == 1){
      return 0;
    }
    return mod-abs(num)%mod;
  }else{
    return num%mod;
  }
}

/*string sentence = And I feel fine...;
    istringstream iss(sentence);
*/  

double eps = .000000001;

int n, m;

int main ()
{
  prep();
  
  cin >> n >> m;
  vector<int> graph[n+1];
  vector<int> rev[n+1];
  for (int i=0; i<m; i++){
      int a, b;
      cin >> a >> b;
      graph[a].push_back(b);
      rev[b].push_back(a);
  }
  int shortest = 1000000000;
  int path[1000];
  for (int i=1; i<=n; i++){
      int dists[n+1];
      bool vis[n+1] = {};
      queue<pair<int, int > > q;
      q.emplace(i, 0);
      bool did = false;
      while (!q.empty()){
          pair<int, int > p = q.front();
          q.pop();
          int node = p.first;
          int dist = p.second;
          if (node == i && dist != 0){
              if (shortest > dist){
                  shortest = dist;
                  did = true;
              }
              break;
          }
          if (vis[node]){
              continue;
          }
          vis[node] = true;
          dists[node] = dist;
          for (int j=0; j<graph[node].size(); j++){
              q.emplace(graph[node][j], dist+1);
          }
      }
      if (!did){
          continue;
      }
      int len = 0;
      // recover path
      int curdist = shortest-1;
      int curnode = i;
      while (curdist >= 0){
          for (int j=0; j<rev[curnode].size(); j++){
              if (dists[rev[curnode][j]] == curdist){
                  curdist--;
                  path[len] = curnode;
                  len++;
                  curnode = rev[curnode][j];
                  break;
              }
          }
      }
  }
  if (shortest == 1000000000){
      cout << -1;
  }else{
      cout << shortest << '\n';
      for (int i=0; i<shortest; i++){
          cout << path[i] << '\n';
      }
  }
  return 0; 
}