#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<int> edge[1005],p,ans;
int vis[1005],now=0,len=1e9;
void dfs(int u,int root,int tag){
    if(tag&&root==u){
        if(len>p.size()){
            len=p.size();
            ans=p;return;
        }
    }
    if(vis[u]==now) return;
    vis[u]=now;
    p.push_back(u);
    for(int i=0;i<edge[u].size();i++){
        dfs(edge[u][i],root,1);
    }
    p.pop_back();
}
int main(){
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=m;i++){
        int u,v;
        cin>>u>>v;
        edge[u].push_back(v);
    }
    for(int i=1;i<=n;i++){
        ++now;
        dfs(i,i,0);
    }
    if(len==1e9){
        cout<<-1<<endl;
        return 0;
    }
    else{
        cout<<len<<endl;
        for(int i=0;i<ans.size();i++){
            cout<<ans[i]<<endl;
        }
    }
}  
### Prompt

In CPP, your task is to solve 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;
typedef long long ll;
vector<int> edge[1005],p,ans;
int vis[1005],now=0,len=1e9;
void dfs(int u,int root,int tag){
    if(tag&&root==u){
        if(len>p.size()){
            len=p.size();
            ans=p;return;
        }
    }
    if(vis[u]==now) return;
    vis[u]=now;
    p.push_back(u);
    for(int i=0;i<edge[u].size();i++){
        dfs(edge[u][i],root,1);
    }
    p.pop_back();
}
int main(){
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=m;i++){
        int u,v;
        cin>>u>>v;
        edge[u].push_back(v);
    }
    for(int i=1;i<=n;i++){
        ++now;
        dfs(i,i,0);
    }
    if(len==1e9){
        cout<<-1<<endl;
        return 0;
    }
    else{
        cout<<len<<endl;
        for(int i=0;i<ans.size();i++){
            cout<<ans[i]<<endl;
        }
    }
}  
```