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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
//#include "atcoder/all"
typedef long long int ll;
using namespace std;
// using namespace atcoder;
#define MAXN 1001
vector<int> vec[MAXN];
bool used[MAXN];
bool tansakutyuu[MAXN];
int cost[MAXN];
int from[MAXN];
bool mark = false;
int check = -1;
void dfs(int x, int dis, int fr) {
    //cout << "x " << x << "fr " << fr << endl;
    if (tansakutyuu[x]) {
        from[x] = fr;
        mark = true;
        check = x;
        return;
    }
    tansakutyuu[x] = true;
    dis++;
    if (cost[x] > dis) {
        from[x] = fr;
        cost[x] = dis;
    }
    for (int i = 0; i < vec[x].size(); i++) {
        dfs(vec[x][i], dis, x);
    }
    tansakutyuu[x] = false;
    used[x] = true;
}
int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
        int a, b;
        cin >> a >> b;
        vec[a].push_back(b);
    }
    memset(used, false, sizeof(used));
    vector<int> ans;
    for (int i = 1; i <= n; i++) {
        if (used[i]) continue;
        for (int j = 0; j < MAXN; j++) {
            cost[j] = MAXN;
        }
        memset(tansakutyuu, false, sizeof(tansakutyuu));
        dfs(i, -1, 0);
        if (mark) {
            int i_st = check;
            check = from[check];
            ans.push_back(i_st);
            while(check != i_st) {
                //cout << check << endl;
                ans.push_back(check);
                check = from[check];
            }
            cout << ans.size() << endl;
            for (int j = ans.size() - 1; j >= 0; j--) {
                cout << ans[j] << endl;
            }
            return 0;
        }
    }
    cout << -1 << endl;
    return 0;
}