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: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <stack>
#include <tuple>
#include <cmath>
#include <iomanip>
#include <map>
#include <cstring> //memset(dp,0,sizeof(dp))
#include <functional>
#define ll long long
#define rep(i,n) for(int i=0;i<(n);i++)
#define repp(i,n) for(int i=n-1;i>=0;i--)
#define fi first
#define se second
#define pb push_back
#define ppb pop_back()
#define ALL(a) (a).begin(),(a).end()
using namespace std;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
typedef pair<int,int> P;
typedef pair<ll,ll> Pll;

signed main(){
    int n,m; cin >> n >> m;
    vector<int> to[n];
    rep(i,m){
        int a,b; cin >> a >> b;
        a--; b--;
        to[a].pb(b);
    }
    bool ok=0;
    vector<int> sta;
    rep(i,n){
        bool bl=0;
        vector<bool> vis(n,0);
        vis[i]=1;
        queue<int> q; q.push(i);
        while(!q.empty()){
            if(bl) break;
            int v=q.front(); q.pop();
            for(int u:to[v]){
                if(u==i){
                    //exist loop
                    sta.pb(i);
                    ok=1; bl=1;
                    break;
                }
                if(vis[u]) continue;
                q.push(u); vis[u]=1;
            }
        }
    }
    if(!ok){
        cout << -1 << endl;
        return 0;
    }
    vector<P> vp;
    vector<int> v[n];
    for(int st:sta){
        //st->stの最短経路を持って復元
        int d[n];
        rep(i,n) d[i]=1001001;
        priority_queue<P,vector<P>,greater<P>> que;
        d[st]=0;
        int okk=0;
        que.push(P(0,st));
        int prev[n];
        while (!que.empty()){
            if(okk==1){
                d[st]=100100;
            }
            okk++;
            P p = que.top();
            que.pop();
            int v = p.second;
            if (d[v]<p.first) continue;
            for (int i=0; i<to[v].size(); ++i) {
                int e = to[v][i];
                if (d[e] > d[v] + 1) {
                    d[e] = d[v] + 1;
                    prev[e]=v;
                    que.push(P(d[e],e));
                }
            }
        }
        vp.pb({d[st],st});
        int stt=st;
        int cnt=d[st];
        rep(i,cnt){
            v[stt].pb(prev[st]);
            st=prev[st];
        }
        reverse(ALL(v[stt]));
    }
    sort(ALL(vp));
    P p=vp[0];
    cout << p.fi << endl;
    for(int u:v[p.se]){
        cout << ++u << endl;
    }
    return 0;
}