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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MAXN=1e3+69;
const int INF = 1e10;
const int mod = 1e9+7;
int m,n,k,q;
int x,y;
int hsh[MAXN];
int Pow[MAXN];
int fact[MAXN];
struct vl{
    int x,y;
    bool operator<(vl t)const{
        if(x==t.x)return y<t.y;
        return x<t.x;
    }
};
int luythua(int x, int y){
    if(y==0)return 1;
    if(y==1)return x%mod;
    int k = luythua(x,y/2);
    if(y%2==0)return k*k%mod;
    return k*k%mod*x%mod;
}
vector<int> a[MAXN];
stack<int> s;
int check[MAXN];
int color[MAXN];
int par[MAXN];
int vao, ra;
int vertex[MAXN];
vector<int> ans;
int pre[MAXN];
int canh[MAXN][MAXN];
bool dfs(int x){
    check[x]=1;
    int ok = 0;
    for(int i=0;i<a[x].size();i++){
        int u = a[x][i];
        if(check[u]==0){
            par[u]=x;
            ok|=dfs(u);
        }
        else if(check[u]==2)continue;
        else{
            ra = x;
            vao = u;
            return true;
        }
    }
    check[x]=2;
    return ok;
}
void ktra(){
    for(int i=0;i<ans.size();i++){
        for(int j=i+1;j<ans.size();j++){
            if(canh[ans[i]][ans[j]]&&pre[ans[j]]!=ans[i]){
                pre[ans[j]]=ans[i];
                ans.clear();
                int v = ans[i];
                while(v!=ans[j]){
                    ans.push_back(v);
                    v = pre[v];
                }
                ans.push_back(v);
                goto ss;
            }
        }
    }
    reverse(ans.begin(),ans.end());
    cout<<ans.size()<<'\n';
    for(int i=0;i<ans.size();i++){
        cout<<ans[i]<<'\n';
    }
    exit(0);
    ss:;
    ktra();
}
void timcycle(){
    for(int i=0;i<=n;i++)check[i]=0;
    for(int j=0;j<=n;j++)par[j]=-1;
    ra = -1;
    for(int i=1;i<=n;i++){
        if(!check[i]&&dfs(i))break;
    }
    if(ra==-1){
        cout<<-1;
    }
    else{
        int v = ra;
        while(v!=vao){
            vertex[v]=1;
            pre[v]=par[v];
            ans.push_back(v);
            v = par[v];
        }
        ans.push_back(v);
        pre[vao]=ra;
        vertex[v]=1;
        ktra();
    }
}
signed main(){                                                                                                                                                                                                                                                                                                  ios_base::sync_with_stdio(false);    cin.tie(0);    cout.tie(0);
    cin>>n>>m;
    for(int i=1;i<=m;i++){
        cin>>x>>y;
        canh[x][y]=1;
        a[x].push_back(y);
    }
    timcycle();
}