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;

#define endl '\n'
#define ALL(a)  (a).begin(),(a).end()
#define ALLR(a)  (a).rbegin(),(a).rend()
#define spa << " " <<
#define lfs <<fixed<<setprecision(10)<<
#define test cout<<"test"<<endl;
#define fi first
#define se second
#define MP make_pair
#define PB push_back
#define EB emplace_back
#define rep(i,n,m) for(ll i = n; i < (ll)(m); i++)
#define rrep(i,n,m) for(ll i = n - 1; i >= (ll)(m); i--)
using ll = long long;
using ld = long double;
const ll MOD = 1e9+7;
//const ll MOD = 998244353;
const ll INF = 1e18;
using P = pair<ll, ll>;
template<typename T>
void chmin(T &a,T b){if(a>b)a=b;}
template<typename T>
void chmax(T &a,T b){if(a<b)a=b;}
void pmod(ll &a,ll b){a=(a+b)%MOD;}
void pmod(ll &a,ll b,ll c){a=(b+c)%MOD;}
void qmod(ll &a,ll b){a=(a*b)%MOD;}
void qmod(ll &a,ll b,ll c){a=(b*c)%MOD;}
ll median(ll a,ll b, ll c){return a+b+c-max({a,b,c})-min({a,b,c});}
void ans1(bool x){if(x) cout<<"Yes"<<endl;else cout<<"No"<<endl;}
void ans2(bool x){if(x) cout<<"YES"<<endl;else cout<<"NO"<<endl;}
void ans3(bool x){if(x) cout<<"Yay!"<<endl;else cout<<":("<<endl;}
template<typename T1,typename T2>
void ans(bool x,T1 y,T2 z){if(x)cout<<y<<endl;else cout<<z<<endl;}  
template<typename T>
void debug(vector<vector<T>>v,ll h,ll w){for(ll i=0;i<h;i++)
{cout<<v[i][0];for(ll j=1;j<w;j++)cout spa v[i][j];cout<<endl;}};
void debug(vector<string>v,ll h,ll w){for(ll i=0;i<h;i++)
{for(ll j=0;j<w;j++)cout<<v[i][j];cout<<endl;}};
template<typename T>
void debug(vector<T>v,ll n){if(n!=0)cout<<v[0];
for(ll i=1;i<n;i++)cout spa v[i];cout<<endl;};
template<typename T>
vector<vector<T>>vec(ll x, ll y, T w){
  vector<vector<T>>v(x,vector<T>(y,w));return v;}
ll gcd(ll x,ll y){ll r;while(y!=0&&(r=x%y)!=0){x=y;y=r;}return y==0?x:y;}
template<typename T>
vector<ll>dx={1,0,-1,0,1,1,-1,-1};
vector<ll>dy={0,1,0,-1,1,-1,1,-1};
template<typename T>
vector<T> make_v(size_t a,T b){return vector<T>(a,b);}
template<typename... Ts>
auto make_v(size_t a,Ts... ts){
  return vector<decltype(make_v(ts...))>(a,make_v(ts...));
}

ll MAX_V = 2e5 + 5;
vector<vector<ll>> G(MAX_V);//隣接リスト(添え字は始点、値は終点)

//引数は各頂点(0-indexed)の入次数
vector<ll>tsort(vector<ll>v){
  ll n=v.size();
  vector<ll>ret(n);
  stack<ll>st;
  for(ll i=0;i<n;i++)if(v[i]==0)st.push(i);
  for(ll i=0;i<n;i++){
    if(st.empty())return vector<ll>();//閉路がある
    ll x=st.top();
    st.pop();
    ret[x]=i;
    for(ll j=0;j<G[x].size();j++){
      v[G[x][j]]--;
      if(v[G[x][j]]==0)st.push(G[x][j]);
    }
  }
  return ret;
}

int main(){
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  ll res=0,res1=INF,res2=-INF,buf=0;
  bool judge = true;
  ll n,m;cin>>n>>m;
  vector<ll>num(n);
  rep(i,0,n){
    ll u,v;cin>>u>>v;
    G[u-1].PB(v-1);
    num[v-1]++;
  }
  auto v=tsort(num);
  ans(v.size()==0,0,-1);
  return 0;
}