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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

struct Edge {
    int to;
};
using Graph = vector<vector<Edge>>;

const int INF = 1e8;
// 幅優先探索
vector<int> dist;  // 始点からの距離を記録
vector<int> pre;
void bfs(const Graph &G, int s) {
    dist.assign((int)G.size(), INF);  // 初期化
    pre.assign((int)G.size(), -1);
    queue<int> que;
    que.emplace(s);  // sから探索する
    dist[s] = 0;
    while (que.size() != 0) {   // 幅優先探索
        int now = que.front();  // 現在の状態
        que.pop();
        for (auto e : G[now]) {
            if (dist[e.to] == INF) {  // 未探索の時のみ行う
                dist[e.to] = dist[now] + 1;
                pre[e.to] = now;
                que.emplace(e.to);
            }
        }
    }
}
/* get_path(prev, t)
    入力：dijkstra で得た prev, ゴール t
    出力： t への最短路のパス
*/
vector<int> get_path(const vector<int> &prev, int t) {
    vector<int> path;
    for (int cur = t; cur != -1; cur = prev[cur]) {
        path.push_back(cur);
    }
    reverse(path.begin(), path.end());
    return path;
}

template <class T>
bool chmin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return 1;
    }
    return 0;
}

int main() {
    // cin.tie(0);
    // ios::sync_with_stdio(false);

    int V, E;
    cin >> V >> E;

    Graph G(V);
    for (int i = 0; i < E; i++) {
        int a, b;
        cin >> a >> b;
        a--, b--;
        G[a].push_back({b});
        // G[b].push_back({a});
    }

    int ans = INF;
    int ans_v = 0;
    int ans_i = 0;
    for (int v = 0; v < V; v++) {
        for (int i = 0; i < (int)G[v].size(); i++) {
            //cout << "ans: " << ans << endl;
            Edge e = G[v][i];
            G[v].erase(G[v].begin() + i);
            bfs(G, e.to);
            if (chmin(ans, dist[v] + 1)) {
                ans_v = v;
                ans_i = i;
            }
            G[v].insert((G[v].begin() + i), e);
        }
    }

    Edge e = G[ans_v][ans_i];
    G[ans_v].erase(G[ans_v].begin() + ans_i);
    bfs(G, e.to);
    if (dist[ans_v] == INF) {
        cout << -1 << endl;
        return 0;
    }
    cout << dist[ans_v] + 1 << "\n";
    for (auto i : get_path(pre, ans_v)) {
        cout << i + 1 << "\n";
    }
    return 0;
}