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;

const int N = 1010;

int n, m, dd;
int depth[N];
bool visited[N];
vector<int> cycle;
vector<int> graph[N];

bool dfs(int u)
{
    int d = -1;
    for (auto v : graph[u])
        d = max(d, depth[v]);

    if (d == -1)
    {
        for (auto v : graph[u])
        {
            depth[v] = depth[u] + 1;
            if (dfs(v))
            {
                if (depth[u] >= dd)
                    cycle.push_back(u);
                return true;
            }
            depth[v] = -1;
        }
    }
    else
    {
        dd = d;
        cycle.push_back(u);
        return true;
    }

    return false;
}

int main()
{
    cin >> n >> m;
    int a, b;
    for (int i = 0; i < m; i++)
        cin >> a >> b,
        graph[a].push_back(b);

    memset(depth, -1, sizeof depth);
    for (int i = 1; i <= n; i++)
    {
        depth[i] = 0;
        if (dfs(i))
        {
            cout << cycle.size() << endl;
            for (auto x : cycle)
                cout << x << endl;
            return 0;
        }
        depth[i] = -1;
    }
    cout << -1;
}