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;

class Node {
  public:
    vector<int> children;
};

bool containsOnlyOne(const unordered_set<int>& s, const vector<int>& v) {
  int cnt = 0;
  for (int x : v) {
    if (s.find(x) != s.end()) {
      cnt++;
    }
  }
  return cnt == 1;
}

// return start
bool dfs(unordered_set<int>& s, vector<Node>& node, int idx, int first) {
  if (s.find(idx) != s.end()) {
    return idx == first;
  }
  s.insert(idx);
  for (int n : node.at(idx).children) {
    bool ret = dfs(s, node, n, first);
    if (ret) {
      if (!containsOnlyOne(s, node.at(idx).children)) {
        return false;
      }
      return ret;
    }
  }
  s.erase(idx);
  return false;
}

int main()
{
  int nnode, nedge;
  cin >> nnode >> nedge;
  vector<Node> node(nnode);
  for (int i = 0; i < nedge; i++) {
    int from, to;
    cin >> from >> to;
    from--;
    to--;
    node.at(from).children.push_back(to);
  }
  unordered_set<int> res;
  for (int i = 0; i < nnode; i++) {
    dfs(res, node, i, i);
    if (!res.empty()) {
      break;
    }
  }
  if (res.empty()) {
    cout << "-1\n";
  } else {
    cout << res.size() << '\n';
    for (int x : res) {
      cout << x+1 << '\n';
    }
  }
  return 0;
}