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: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <algorithm>
#include <set>
#include <map>
#include <bitset>
#include <cmath>
#include <functional>
#include <iomanip>
#define vll vector<ll>
#define vvv vector<vvl>
#define vvi vector<vector<int> >
#define vvl vector<vector<ll> >
#define vv(a, b, c, d) vector<vector<d> >(a, vector<d>(b, c))
#define vvvl(a, b, c, d) vector<vvl>(a, vvl(b, vll (c, d)));
#define rep(c, a, b) for(ll c=a;c<b;c++)
#define re(c, b) for(ll c=0;c<b;c++)
#define all(obj) (obj).begin(), (obj).end()
typedef long long int ll;
typedef long double ld;
using namespace std;

vvl G = vv(1001, 0, 0, ll);
vll tmp, ans(1001, 0);
vll use(1001, 0), par(1001, 0);

void dfs(ll now, ll st){
  for(int i=0;i<G[now].size();i++){
    ll to = G[now][i];
    if(use[to]&&to!=st) return;
  }
  use[now] = 1;
  for(int i=0;i<G[now].size();i++) {
    ll to = G[now][i];
    if(to==st){
      tmp.clear();
      ll k = now;
      while(k!=-1){
        tmp.push_back(k);
        k = par[k];
      }
      if(ans.size()>tmp.size()) ans = tmp;
    }else{
      par[to] = now;
      dfs(G[now][i], st);
    }
  }
  use[now] = 0;
}
int main(int argc, char const *argv[]) {
  ll a, b, n, m;std::cin >> n >> m;
  re(i, m){
    std::cin >> a >> b;
    G[a].push_back(b);
  }
  for(int i=1;i<=n;i++){
    re(j, n+1) par[j] = -1;
    dfs(i, i);
  }
  if(ans.size()==1001) std::cout << -1 << '\n';
  else{
    std::cout << ans.size() << '\n';
    re(i, ans.size()) std::cout << ans[ans.size()-1-i] << '\n';
  }
  return 0;
}