#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MOD = 1000000007;
vector<int> k[1000],ans;
bool t[1000*1000]={};
 int used[1000]={},c=-1;
void dfs(int a){
  
  if(used[a]==1)return ;
  if(used[a]==-1){
    c=a;
    return ;
  }
  used[a]=-1;
  for(int i=0;i<k[a].size();i++){
    dfs(k[a][i]);
    if(c!=-1)break;
  }
  if(c>=0){
   
    ans.push_back(a+1);
    if(a==c)c=-2;
   
  }
  used[a]=1;
  return ;
}
signed main() {
  int n,m;
  cin>>n>>m;
  int a,b;
 
  for(int i=0;i<m;i++){
    cin>>a>>b;
    k[a-1].push_back(b-1);
    t[(a-1)*n+b-1]=true;
  }
 
  stack<int> s;
  for(int i=0;i<n;i++){
   
    dfs(i);
    if(c!=-1)break;
  }
 if(ans.empty()){
   cout<<-1;
   return 0;
 }
 reverse(ans.begin(),ans.end());

  int u=0,w;
  while(ans[u]>=0){
    w=ans.size()-1;
    while(!t[(ans[u]-1)*n+abs(ans[w])-1])w--;
    ans.push_back(-ans[u]);
    u=w;
  }
  cout<<ans.size()-u<<endl;
  for(int i=u;i<ans.size();i++)cout<<-ans[i]<<endl;

  
  


  


  return 0;
}

### Prompt

Create a solution in cpp for 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 int long long
const int MOD = 1000000007;
vector<int> k[1000],ans;
bool t[1000*1000]={};
 int used[1000]={},c=-1;
void dfs(int a){
  
  if(used[a]==1)return ;
  if(used[a]==-1){
    c=a;
    return ;
  }
  used[a]=-1;
  for(int i=0;i<k[a].size();i++){
    dfs(k[a][i]);
    if(c!=-1)break;
  }
  if(c>=0){
   
    ans.push_back(a+1);
    if(a==c)c=-2;
   
  }
  used[a]=1;
  return ;
}
signed main() {
  int n,m;
  cin>>n>>m;
  int a,b;
 
  for(int i=0;i<m;i++){
    cin>>a>>b;
    k[a-1].push_back(b-1);
    t[(a-1)*n+b-1]=true;
  }
 
  stack<int> s;
  for(int i=0;i<n;i++){
   
    dfs(i);
    if(c!=-1)break;
  }
 if(ans.empty()){
   cout<<-1;
   return 0;
 }
 reverse(ans.begin(),ans.end());

  int u=0,w;
  while(ans[u]>=0){
    w=ans.size()-1;
    while(!t[(ans[u]-1)*n+abs(ans[w])-1])w--;
    ans.push_back(-ans[u]);
    u=w;
  }
  cout<<ans.size()-u<<endl;
  for(int i=u;i<ans.size();i++)cout<<-ans[i]<<endl;

  
  


  


  return 0;
}

```