#include <bits/stdc++.h>
using namespace std;
const int MOD=1e9+7;
//const int MOD=998244353;
const int INF=1e9;
const long long LINF=1e18;
#define int long long
//template
template <typename T>
void fin(T a){
  cout<<a<<endl;
  exit(0);
}
//ありがとうありがとうありがとうありがとうありがとうありがとうありがとうありがとうありがとうありがとう
//ありがとうありがとうありがとうありがとうありがとうありがとうありがとうありがとうありがとうありがとう
std::vector<int> used(10000,0),ans,fans,edge[10000];
bool FIN=false;

int dfs(int now,int cnt){
  if(FIN)return 1;

  used[now]=cnt;

  int t=0,w;
  for(int p:edge[now])
    if(used[p]>t)
      t=used[p],w=p;

  if(t){
    ans.push_back(now);
    used[w]=-1;
    return 1;
  }

  for(int p:edge[now]){
    if(FIN)return 1;
    if(dfs(p,cnt+1)){
      if(FIN)return 1;
      ans.push_back(now);
      if(used[now]<0)FIN=true;
      return 1;
    }
  }

  used[now]=0;
  return 0;
}

//main
signed main(){
  int N,M;cin>>N>>M;

  while(M--){
    int a,b;cin>>a>>b;a--;b--;
    edge[a].push_back(b);
  }

  for(int i=0;i<N;i++){
    FIN=0;
    fill(used.begin(),used.end(),0);
    if(dfs(i,1)){
      if(ans.size()<fans.size()||!fans.size())fans=ans;
      ans.clear();
    }
  }


  if(fans.size()){
    cout<<fans.size()<<endl;
    for(int p:fans)cout<<p+1<<endl;
  }
  else fin(-1);
}

### Prompt

Create a solution in cpp for the following problem:

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.

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.

Examples

Input

4 5
1 2
2 3
2 4
4 1
4 3


Output

3
1
2
4


Input

4 5
1 2
2 3
2 4
1 4
4 3


Output

-1


Input

6 9
1 2
2 3
3 4
4 5
5 6
5 1
5 2
6 1
6 2


Output

4
2
3
4
5

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD=1e9+7;
//const int MOD=998244353;
const int INF=1e9;
const long long LINF=1e18;
#define int long long
//template
template <typename T>
void fin(T a){
  cout<<a<<endl;
  exit(0);
}
//ありがとうありがとうありがとうありがとうありがとうありがとうありがとうありがとうありがとうありがとう
//ありがとうありがとうありがとうありがとうありがとうありがとうありがとうありがとうありがとうありがとう
std::vector<int> used(10000,0),ans,fans,edge[10000];
bool FIN=false;

int dfs(int now,int cnt){
  if(FIN)return 1;

  used[now]=cnt;

  int t=0,w;
  for(int p:edge[now])
    if(used[p]>t)
      t=used[p],w=p;

  if(t){
    ans.push_back(now);
    used[w]=-1;
    return 1;
  }

  for(int p:edge[now]){
    if(FIN)return 1;
    if(dfs(p,cnt+1)){
      if(FIN)return 1;
      ans.push_back(now);
      if(used[now]<0)FIN=true;
      return 1;
    }
  }

  used[now]=0;
  return 0;
}

//main
signed main(){
  int N,M;cin>>N>>M;

  while(M--){
    int a,b;cin>>a>>b;a--;b--;
    edge[a].push_back(b);
  }

  for(int i=0;i<N;i++){
    FIN=0;
    fill(used.begin(),used.end(),0);
    if(dfs(i,1)){
      if(ans.size()<fans.size()||!fans.size())fans=ans;
      ans.clear();
    }
  }


  if(fans.size()){
    cout<<fans.size()<<endl;
    for(int p:fans)cout<<p+1<<endl;
  }
  else fin(-1);
}

```