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 NMAX = 1005;
int n,m,fathr[NMAX],dist[NMAX], sz;
vector < int > v[NMAX];
vector < int > vans;
vector < int > work;
int solution;

void antidfs(int node, int stop){
    work.clear();
    while(node != stop){
        work.push_back(node);
        node = fathr[node];
    }
    work.push_back(stop);
    if(work.size() < sz){
        sz = work.size();
        vans = work;
    }
}

void dfs(int node, int father){
    if(solution)
        return;
    for(auto it: v[node])
        if(it != father && !solution){
            if(!dist[it]){
                dist[it] = dist[node] + 1;
                fathr[it] = node;
                dfs(it,father);
            }else
                if(dist[it] < dist[node]){
                    antidfs(node, it);
                    antidfs(node, it);
                    antidfs(node, it);
                    antidfs(node, it);
                    solution = 1;
                }
        }
}

int main(){
    int i,j,x,y;
    cin >> n >> m;
    for(i = 1 ; i <= m ; i++){
        cin >> x >> y;
        v[x].push_back(y);
    }

    sz = 2 * NMAX;
    dfs(1,0);

    if(vans.size() == 0){
        cout << -1;
        return 0;
    }

    cout << vans.size() << "\n";
    for(auto it: vans)
        cout << it << "\n";

    return 0;
}