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;

int n,m;
vector<int> G[1010];
int d[1010];
int pre[1010];

int main() {
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int from, to;
        cin >> from >> to;
        G[from].push_back(to);
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) d[j] = (1<<19);
        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> pq;
        pq.push({0,i});
        d[i] = 0;
        while (!pq.empty()) {
            pair<int,int> tp = pq.top();
            pq.pop();
            if (tp.first > d[tp.second]) continue;
            for (int j = 0; j < G[tp.second].size(); j++) {
                int to = G[tp.second][j];
                if (to == i) {

                    stack<int> stk;
                    int now = tp.second;
                    while (now != i) {
                        stk.push(now);
                        now = pre[now];
                    }
                    stk.push(i);

                    cout << stk.size() << endl;
                    while (!stk.empty()) {
                        cout << stk.top() << endl;
                        stk.pop();
                    }

                    return 0;

                }
                if (d[to] > tp.first + 1) {
                    d[to] = tp.first + 1;
                    pq.push({d[to], to});
                    pre[to] = tp.second;
                }
            }
        }

    }
    cout << -1 << endl;
    return 0;
}