#include <bits/stdc++.h>

using namespace std;

vector<int>li[1001];
bool vis[1001];
bool rec[1001];

int par[1001];

int n,m;

vector<pair<int,vector<int> > >possible;

void dfs(int v){

    vis[v]=true;
    rec[v]=true;

    for(auto e:li[v]){
        if(rec[e]){
            
            
            vector<int>ans;

            ans.push_back(e);

            int curr=v;
            
            while(curr!=e){
                
                ans.push_back(curr);
                curr=par[curr];
            }

            

            possible.push_back({(int)ans.size(),ans});

        }

        if(!vis[e]){
            par[e]=v;
            dfs(e);
        }
    }

    rec[v]=false;
}

int main() {
   
   cin>>n>>m;
   
   for(int i=0;i<m;i++){
        int a,b;
        cin>>a>>b;

        li[a].push_back(b);
   
   }
   
   for(int i=1;i<=n;i++)
    sort(li[i].begin(),li[i].end());

   for(int i=1;i<=n;i++){

        if(!vis[i]){
            
            par[i]=-1;
            
            dfs(i);

        }

   }
    
    
  
   if(possible.empty())
   cout<<-1;
   else{
       sort(possible.begin(),possible.end());
       
       cout<<possible[0].first<<"\n";
       
       for(auto e:possible[0].second){
           cout<<e<<"\n";
       }
   }

   
}
### 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;

vector<int>li[1001];
bool vis[1001];
bool rec[1001];

int par[1001];

int n,m;

vector<pair<int,vector<int> > >possible;

void dfs(int v){

    vis[v]=true;
    rec[v]=true;

    for(auto e:li[v]){
        if(rec[e]){
            
            
            vector<int>ans;

            ans.push_back(e);

            int curr=v;
            
            while(curr!=e){
                
                ans.push_back(curr);
                curr=par[curr];
            }

            

            possible.push_back({(int)ans.size(),ans});

        }

        if(!vis[e]){
            par[e]=v;
            dfs(e);
        }
    }

    rec[v]=false;
}

int main() {
   
   cin>>n>>m;
   
   for(int i=0;i<m;i++){
        int a,b;
        cin>>a>>b;

        li[a].push_back(b);
   
   }
   
   for(int i=1;i<=n;i++)
    sort(li[i].begin(),li[i].end());

   for(int i=1;i<=n;i++){

        if(!vis[i]){
            
            par[i]=-1;
            
            dfs(i);

        }

   }
    
    
  
   if(possible.empty())
   cout<<-1;
   else{
       sort(possible.begin(),possible.end());
       
       cout<<possible[0].first<<"\n";
       
       for(auto e:possible[0].second){
           cout<<e<<"\n";
       }
   }

   
}
```