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;

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

  int N, M; cin>>N>>M;
  vector<vector<int>> graph(N);
  for (int i = 0; i < M; i++) {
	int a, b; cin>>a>>b;
	a--; b--;
	graph[a].push_back(b);
  }
  auto find_circuit = [f=[&](auto g, int root, int c, set<int>& loute) -> set<int> {
	for (auto t : graph[c]) {
	  if (loute.count(t))
	    return loute;
	  loute.insert(t);
	  auto res = g(g, root, t, loute);
	  if (not res.empty()) {
	    return res;
	  }
	  loute.erase(t);
	}
	return {};
  }](int root) {
    set<int> loute{root};
	return f(f, root, root, loute);
  };

  auto make_graph = [&](set<int> loute) {
	vector<vector<int>> ngraph(N);
	for (auto s : loute) {
	  for (auto t : graph[s]) {
		if (loute.count(t)) {
		  ngraph[s].push_back(t);
		}
	  }
	}
	return ngraph;
  };
  set<int> loute;
  for (int i = 0; i < N; i++) {
	loute = find_circuit(i);
	if (not loute.empty()) {
	  graph = make_graph(loute);
	  break;
	}
  }
  if (loute.empty()) {
    cout << -1 << endl;
    return 0;
  }

  set<int> ans;
  while (true) {
    map<int, int> in;
    for (auto& s : graph) {
      for (auto t : s)
        in[t]++;
    }
    int c = -1;
    for (auto i : in) {
      if (i.second >= 2) {
		c = i.first;
		break;
      }
    }
	if (c == -1) {
	  ans = loute;
	  break;
	}
	
	int t = graph[c].front();
	loute = {t};
	while (t != c) {
	  t = graph[c].front();
	  loute.insert(t);
	}
	graph = make_graph(loute);
  }
  cout << ans.size() << endl;
  for (auto s : ans)
	cout << s+1 << endl;

  return 0;
}