#include <bits/stdc++.h>
using namespace std;
using ll=long long;
#define MOD 1000000007
#define MAX 2100000

int main(){
  int N,M;
  cin>>N>>M;
  vector<int> A(M),B(M);
  vector<vector<int>> G(N),H(N);
  for(int i=0;i<M;i++){
    cin>>A.at(i)>>B.at(i);
    A.at(i)--,B.at(i)--;
    G.at(A.at(i)).push_back(B.at(i));
    H.at(B.at(i)).push_back(A.at(i));
  }

  vector<int> Ans(10000);
  for(int i=0;i<N;i++){
    vector<int> dist(N,100000);
    dist.at(i)=0;
    queue<int> q;
    q.push(i);
    int y=-1;
    while(!q.empty()){
      int v=q.front();
      for(auto nv:G.at(v)){
        if(nv==i){
          y=v;
          break;
        }
        if(dist.at(nv)==100000){
          dist.at(nv)=dist.at(v)+1;
          q.push(nv);
        }
      }
      q.pop();
    }
    if(y==-1){
      continue;
    }
    vector<int> ans;
    ans.push_back(y);
    while(y!=i){
      for(auto ny:H.at(y)){
        if(dist.at(y)==dist.at(ny)+1){
          y=ny;
          break;
        }
      }
      ans.push_back(y);
    }
    if(ans.size()<Ans.size()){
      Ans=ans;
    }
  }

  if(Ans.size()==10000){
    cout<<-1<<endl;
  }else{
    cout<<Ans.size()<<endl;
    for(int i=0;i<Ans.size();i++){
      cout<<Ans.at(i)+1<<endl;
    }
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
using ll=long long;
#define MOD 1000000007
#define MAX 2100000

int main(){
  int N,M;
  cin>>N>>M;
  vector<int> A(M),B(M);
  vector<vector<int>> G(N),H(N);
  for(int i=0;i<M;i++){
    cin>>A.at(i)>>B.at(i);
    A.at(i)--,B.at(i)--;
    G.at(A.at(i)).push_back(B.at(i));
    H.at(B.at(i)).push_back(A.at(i));
  }

  vector<int> Ans(10000);
  for(int i=0;i<N;i++){
    vector<int> dist(N,100000);
    dist.at(i)=0;
    queue<int> q;
    q.push(i);
    int y=-1;
    while(!q.empty()){
      int v=q.front();
      for(auto nv:G.at(v)){
        if(nv==i){
          y=v;
          break;
        }
        if(dist.at(nv)==100000){
          dist.at(nv)=dist.at(v)+1;
          q.push(nv);
        }
      }
      q.pop();
    }
    if(y==-1){
      continue;
    }
    vector<int> ans;
    ans.push_back(y);
    while(y!=i){
      for(auto ny:H.at(y)){
        if(dist.at(y)==dist.at(ny)+1){
          y=ny;
          break;
        }
      }
      ans.push_back(y);
    }
    if(ans.size()<Ans.size()){
      Ans=ans;
    }
  }

  if(Ans.size()==10000){
    cout<<-1<<endl;
  }else{
    cout<<Ans.size()<<endl;
    for(int i=0;i<Ans.size();i++){
      cout<<Ans.at(i)+1<<endl;
    }
  }
}

```