#include <bits/stdc++.h>

using namespace std;
#define ll long long
const int N=2005,INF=1e9;
bool vis[N],stk[N];
int in[N];
stack<int>st;
vector<int>adj[N];
int t=0;
vector<int>ans;
void solve(int u)
{
    ans.push_back(u);
    int mx=-1,e;
    for(auto v:adj[u])
        if(stk[v])
        {
            if(in[v]>mx)
            {
                mx=in[v];
                e=v;
            }
        }
    if(e!=ans[0])solve(e);
}
void dfs(int u)
{
    vis[u]=1;
    stk[u]=1;
    in[u]=t++;
    st.push(u);
    int mx=-1,e;
    for(auto v:adj[u])
        if(stk[v])
        {
            if(in[v]>mx)
            {
                mx=in[v];
                e=v;
            }
        }
    if(mx!=-1)
    {
        solve(e);
        cout<<ans.size()<<'\n';
        for(auto x:ans)
            cout<<x<<'\n';
        exit(0);

    }
    for(auto v:adj[u])
        if(!vis[v])
            dfs(v);

    stk[u]=0;
    st.pop();
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    int n,m;
    cin>>n>>m;
    for(int i=0;i<m;i++)
    {
        int u,v;
        cin>>u>>v;
        adj[u].push_back(v);
    }
    for(int i=1;i<=n;i++)
    {
        memset(vis,0,sizeof vis);
        t=0;
        dfs(i);
    }
    cout<<-1<<endl;
    return 0;
}

### Prompt

Please formulate 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;
#define ll long long
const int N=2005,INF=1e9;
bool vis[N],stk[N];
int in[N];
stack<int>st;
vector<int>adj[N];
int t=0;
vector<int>ans;
void solve(int u)
{
    ans.push_back(u);
    int mx=-1,e;
    for(auto v:adj[u])
        if(stk[v])
        {
            if(in[v]>mx)
            {
                mx=in[v];
                e=v;
            }
        }
    if(e!=ans[0])solve(e);
}
void dfs(int u)
{
    vis[u]=1;
    stk[u]=1;
    in[u]=t++;
    st.push(u);
    int mx=-1,e;
    for(auto v:adj[u])
        if(stk[v])
        {
            if(in[v]>mx)
            {
                mx=in[v];
                e=v;
            }
        }
    if(mx!=-1)
    {
        solve(e);
        cout<<ans.size()<<'\n';
        for(auto x:ans)
            cout<<x<<'\n';
        exit(0);

    }
    for(auto v:adj[u])
        if(!vis[v])
            dfs(v);

    stk[u]=0;
    st.pop();
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    int n,m;
    cin>>n>>m;
    for(int i=0;i<m;i++)
    {
        int u,v;
        cin>>u>>v;
        adj[u].push_back(v);
    }
    for(int i=1;i<=n;i++)
    {
        memset(vis,0,sizeof vis);
        t=0;
        dfs(i);
    }
    cout<<-1<<endl;
    return 0;
}

```