#include<bits/stdc++.h>
#define long long long
using namespace std;
const int N=1e6;
vector<int>adj[N+2];
vector<pair<int,int> >bk;
int vis[1000+2],lev[N+2],prnt[1000+2];
void dfs(int node,int par,int lv)
{
  lev[node]=lv;prnt[node]=par;
  vis[node]=1;
  for(auto x:adj[node])
  {
    if(vis[x])bk.push_back({node,x});
    else dfs(x,node,lv+1);
  }
}
int main()
{
  //freopen("inp.txt","r",stdin);
  ios_base::sync_with_stdio(0);cin.tie(0);
  int n,m;cin>>n>>m;
  for(int i=1;i<=n;i++)adj[i].clear();
  for(int i=1;i<=m;i++)
  {
    int u,v;cin>>u>>v;
    adj[u].push_back(v);
  }
  int mn=n+1,u,v;vector<int>pr;
  for(int i=1;i<=n;i++)
  {
    memset(vis,0,sizeof(vis));
    bk.clear();dfs(i,-1,1);
    for(auto x:bk)
    {
      int df=lev[x.first]-lev[x.second];
      if(df>0&&df<mn)
      {
        int f=0;u=x.first,v=x.second;vector<int>tmp;
        while(true)
        {
          if(u==-1){f=1;break;}
          tmp.push_back(u);if(u==v)break;
          u=prnt[u];
        }
        if(!f)mn=df,pr=tmp;
      }
    }
  }
  if(mn>n)cout<<-1<<endl,exit(0);
  cout<<mn+1<<"\n";
  for(auto x:pr)
    cout<<x<<"\n";
  return 0;
}

### 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>
#define long long long
using namespace std;
const int N=1e6;
vector<int>adj[N+2];
vector<pair<int,int> >bk;
int vis[1000+2],lev[N+2],prnt[1000+2];
void dfs(int node,int par,int lv)
{
  lev[node]=lv;prnt[node]=par;
  vis[node]=1;
  for(auto x:adj[node])
  {
    if(vis[x])bk.push_back({node,x});
    else dfs(x,node,lv+1);
  }
}
int main()
{
  //freopen("inp.txt","r",stdin);
  ios_base::sync_with_stdio(0);cin.tie(0);
  int n,m;cin>>n>>m;
  for(int i=1;i<=n;i++)adj[i].clear();
  for(int i=1;i<=m;i++)
  {
    int u,v;cin>>u>>v;
    adj[u].push_back(v);
  }
  int mn=n+1,u,v;vector<int>pr;
  for(int i=1;i<=n;i++)
  {
    memset(vis,0,sizeof(vis));
    bk.clear();dfs(i,-1,1);
    for(auto x:bk)
    {
      int df=lev[x.first]-lev[x.second];
      if(df>0&&df<mn)
      {
        int f=0;u=x.first,v=x.second;vector<int>tmp;
        while(true)
        {
          if(u==-1){f=1;break;}
          tmp.push_back(u);if(u==v)break;
          u=prnt[u];
        }
        if(!f)mn=df,pr=tmp;
      }
    }
  }
  if(mn>n)cout<<-1<<endl,exit(0);
  cout<<mn+1<<"\n";
  for(auto x:pr)
    cout<<x<<"\n";
  return 0;
}

```