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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
#define MOD 1000000007
#define MAX 2100000

set<int> ans;
bool dfs(vector<vector<int>> &G,int v,vector<bool> visit){
  visit.at(v)=true;
  for(auto nv: G.at(v)){
    if(visit.at(nv)){
      ans.insert(v);
      return true;
    }
    bool flag=dfs(G,nv,visit);
    if(flag){
      ans.insert(v);
      return true;
    }
  }
  return false;
}

int main(){
  int N,M;
  cin>>N>>M;
  vector<int> A(M),B(M);
  vector<vector<int>> G(N),H(N);
  for(int i=0;i<M;i++){
    cin>>A.at(i)>>B.at(i);
    A.at(i)--,B.at(i)--;
    G.at(A.at(i)).push_back(B.at(i));
    H.at(B.at(i)).push_back(A.at(i));
  }

  bool flag=false;
  for(int i=0;i<N;i++){
    vector<bool> visit(N,false);
    flag=dfs(G,i,visit);
    if(flag){
      break;
    }
  }

  if(flag){
    int n=ans.size();
    vector<int> ANS(1000);
    for(auto x:ans){
      vector<int> dist(N,100000);
      dist.at(x)=0;
      queue<int> q;
      q.push(x);
      int y=-1;
      while(!q.empty()){
        int v=q.front();
        for(auto nv:G.at(v)){
          if(nv==x){
            y=v;
            break;
          }
          if(ans.count(nv)==1){
            if(dist.at(nv)==100000){
              dist.at(nv)=dist.at(v)+1;
              q.push(nv);
            }
          }
        }
        if(y!=-1){
          break;
        }
        q.pop();
      }
      vector<int> Ans;
      Ans.push_back(y);
      while(y!=x){
        for(auto ny:H.at(y)){
          if(dist.at(y)==dist.at(ny)+1){
            y=ny;
            break;
          }
        }
        Ans.push_back(y);
      }
      if(Ans.size()<ANS.size()){
        ANS=Ans;
      }
    }
    cout<<ANS.size()<<endl;
    for(int i=0;i<ANS.size();i++){
      cout<<ANS.at(i)+1<<endl;
    }
  }else{
    cout<<-1<<endl;
  }
}