#include<bits/stdc++.h>
using namespace std;
const int mxN=1e3+10;
int n,m,a,b,k;
int vis[mxN],pai[mxN];
vector<int>adj[mxN];
vector<int>resp;

void bfs(int x){
    queue<int>fila;
    fila.push(x);
    vis[x]=1;

    while(!fila.empty()){
        int u=fila.front();
        fila.pop();
        for(int v:adj[u])   {
            if(!vis[v]){
                vis[v]=1;
                pai[v]=u;
                fila.push(v);

            }
            if(v==x){
                vector<int>ans;
                while(u!=v){
                    ans.push_back(u);
                    u=pai[u];
                }
                ans.push_back(v);
                if(resp.size()==0 || ans.size()< resp.size())resp=ans;
                return;
            }
        }

    }
}

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    cin>>n>>m;
    for(int i=0;i<m;i++){
        cin>>a>>b;
        adj[a].push_back(b);
    }

    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            pai[j]=-1;
            vis[j]=0;            
        }
        bfs(i);       
    }
    if(resp.size()){
        cout<<resp.size()<<"\n";
        for(auto x:resp){
            cout<<x<<"\n";
        }
        return 0;
    }
    cout<<-1<<"\n";

     

}
### Prompt

Your challenge is to write a Cpp solution to 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 mxN=1e3+10;
int n,m,a,b,k;
int vis[mxN],pai[mxN];
vector<int>adj[mxN];
vector<int>resp;

void bfs(int x){
    queue<int>fila;
    fila.push(x);
    vis[x]=1;

    while(!fila.empty()){
        int u=fila.front();
        fila.pop();
        for(int v:adj[u])   {
            if(!vis[v]){
                vis[v]=1;
                pai[v]=u;
                fila.push(v);

            }
            if(v==x){
                vector<int>ans;
                while(u!=v){
                    ans.push_back(u);
                    u=pai[u];
                }
                ans.push_back(v);
                if(resp.size()==0 || ans.size()< resp.size())resp=ans;
                return;
            }
        }

    }
}

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    cin>>n>>m;
    for(int i=0;i<m;i++){
        cin>>a>>b;
        adj[a].push_back(b);
    }

    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            pai[j]=-1;
            vis[j]=0;            
        }
        bfs(i);       
    }
    if(resp.size()){
        cout<<resp.size()<<"\n";
        for(auto x:resp){
            cout<<x<<"\n";
        }
        return 0;
    }
    cout<<-1<<"\n";

     

}
```