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)

/*
[abc142] F - Pure
*/

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;
#define ALL(c) (c).begin(), (c).end()

const int MAX_N = 1000;
const int MAX_M = 2000;

int N, M;
int A[MAX_M], B[MAX_M];
vector<int> g[MAX_N];
int visited[MAX_N];

bool dfs(int v, vector<int> &cycle) {
    visited[v] = 1;
    for (int to : g[v]) {
        if (visited[to] == 2) {
            continue;
        }
        if (visited[to] == 1 || dfs(to, cycle)) {
            cycle.push_back(v);
            return true;
        }
    }
    visited[v] = 2;
    return false;
}

pii check(vector<int> &cycle) {
    vector<bool> used(N);
    int cl = cycle.size();
    for (int v : cycle) {
        used[v] = true;
    }
    for (int i = 0; i < cl; i++) {
        int v = cycle[i];
        for (int to : g[v]) {
            if (used[to] && to != cycle[(i + 1) % cl]) {
                return {v, to};
            }
        }
    }
    return {-1, -1};
}

void solve() {
    for (int i = 0; i < M; i++) {
        g[A[i]].push_back(B[i]);
    }

    vector<int> cycle;
    for (int i = 0; i < N; i++) {
        if (!visited[i] && dfs(i, cycle)) {
            reverse(ALL(cycle));
            break;
        }
    }
    if (cycle.empty()) {
        cout << -1 << endl;
        return;
    }
    for (int i = 1; i < (int)cycle.size(); i++) {
        if (cycle[i] == cycle[0]) {
            cycle.resize(i);
            break;
        }
    }

    for (int k = 1; k < N; k++) {
        pii xy = check(cycle);
        if (xy.first == -1) {
            break;
        }

        int cl = cycle.size();
        int i;
        for (i = 0; i < cl; i++) {
            if (cycle[i] == xy.second) {
                break;
            }
        }
        vector<int> nc = {xy.first};
        while (cycle[i] != xy.first) {
            nc.push_back(cycle[i]);
            i = (i + 1) % cl;
        }
        cycle = nc;
    }

    cout << cycle.size() << endl;
    for (int v : cycle) {
        cout << v + 1 << endl;
    }
}

int main() {
    cin >> N >> M;
    for (int i = 0; i < M; i++) {
        cin >> A[i] >> B[i];
        A[i]--, B[i]--;
    }

    solve();

    return 0;
}