#include <bits/stdc++.h>

typedef long long ll;
#define int ll

using namespace std;

const int N=1010;

vector<int> g[N],r[N];
bool used[N];
vector<int> ans;

void dfs(int v)
{
    if(used[v])
        return;
    used[v]=true;
    int mn=int(ans.size())-1;
    for(int from:r[v])
        for(int i=0;i<ans.size();i++)
            if(ans[i]==from)
                mn=min(mn,i);
    vector<int> del(ans.begin()+mn+1,ans.end());
    ans.erase(ans.begin()+mn+1,ans.end());
    ans.push_back(v);
    int mx=-1;
    for(int to:g[v])
        for(int i=0;i<ans.size();i++)
            if(ans[i]==to)
                mx=max(mx,i);
    if(mx>=0)
    {
        cout<<ans.size()-mx<<"\n";
        for(int i=mx;i<ans.size();i++)
            cout<<ans[i]<<"\n";
        exit(0);
    }
    for(int to:g[v])
        dfs(to);
    ans.pop_back();
    ans.insert(ans.end(),del.begin(),del.end());
}

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

    int n,m;
    cin>>n>>m;
    while(m--)
    {
        int a,b;
        cin>>a>>b;
        g[a].push_back(b);
        r[b].push_back(a);
    }
    for(int i=1;i<=n;i++)
        dfs(i);
    cout<<-1;
}

### Prompt

Please create a solution in CPP 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>

typedef long long ll;
#define int ll

using namespace std;

const int N=1010;

vector<int> g[N],r[N];
bool used[N];
vector<int> ans;

void dfs(int v)
{
    if(used[v])
        return;
    used[v]=true;
    int mn=int(ans.size())-1;
    for(int from:r[v])
        for(int i=0;i<ans.size();i++)
            if(ans[i]==from)
                mn=min(mn,i);
    vector<int> del(ans.begin()+mn+1,ans.end());
    ans.erase(ans.begin()+mn+1,ans.end());
    ans.push_back(v);
    int mx=-1;
    for(int to:g[v])
        for(int i=0;i<ans.size();i++)
            if(ans[i]==to)
                mx=max(mx,i);
    if(mx>=0)
    {
        cout<<ans.size()-mx<<"\n";
        for(int i=mx;i<ans.size();i++)
            cout<<ans[i]<<"\n";
        exit(0);
    }
    for(int to:g[v])
        dfs(to);
    ans.pop_back();
    ans.insert(ans.end(),del.begin(),del.end());
}

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

    int n,m;
    cin>>n>>m;
    while(m--)
    {
        int a,b;
        cin>>a>>b;
        g[a].push_back(b);
        r[b].push_back(a);
    }
    for(int i=1;i<=n;i++)
        dfs(i);
    cout<<-1;
}

```