#include <bits/stdc++.h>
#define int long long
#define N 1005
using namespace std;

int n, m;
vector<int> G[N];
vector<int> rG[N];
int used[N], line[N], start, f;

void dfs(int x, int px){
  
  if( f ) return;
  
  used[x] = 1;
  line[x] = 1;
  
  int flg = 0;
  
  for(int nx : G[x] ){
    
    if( nx == start ){
      flg = 1;
    }
    
    if( nx != start && line[nx] ){
      line[x] = 0;
      return;
    }
    
  }
  
  for(int nx : rG[x] ){
    
    if( line[nx] && nx != px ){
      line[x] = 0;
      return;
    }
    
  }
  
  if( flg ){
    f = 1;
    return;
  }
  
  for(int nx : G[x] ){
    
    if( used[nx] ) continue;
    
    dfs( nx, x );

    if( f ) return;
    
  }
  
  line[x] = 0;
  
}

signed main(){
  
  cin>>n>>m;
  
  for(int i=0;i<m;i++){
    int a, b;
    cin>>a>>b;
    G[a-1].push_back(b-1);
    rG[b-1].push_back(a-1);
  }
  
  for(int i=0;i<n;i++){
    start = i;
    memset( used, 0, sizeof(used) );
    dfs(i,-1);
    if( f ) break;
  }
  
  if( !f ){
    cout<<-1<<endl;
    return 0;
  }
  
  int cnt = 0;
  
  for(int i=0;i<n;i++){
    if( line[i] ) cnt++;
  }
  
  cout<<cnt<<endl;
  
  for(int i=0;i<n;i++){
    if( line[i] ) cout<<i+1<<endl;
  }
  
  return 0;
}

### Prompt

Your task is to create 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>
#define int long long
#define N 1005
using namespace std;

int n, m;
vector<int> G[N];
vector<int> rG[N];
int used[N], line[N], start, f;

void dfs(int x, int px){
  
  if( f ) return;
  
  used[x] = 1;
  line[x] = 1;
  
  int flg = 0;
  
  for(int nx : G[x] ){
    
    if( nx == start ){
      flg = 1;
    }
    
    if( nx != start && line[nx] ){
      line[x] = 0;
      return;
    }
    
  }
  
  for(int nx : rG[x] ){
    
    if( line[nx] && nx != px ){
      line[x] = 0;
      return;
    }
    
  }
  
  if( flg ){
    f = 1;
    return;
  }
  
  for(int nx : G[x] ){
    
    if( used[nx] ) continue;
    
    dfs( nx, x );

    if( f ) return;
    
  }
  
  line[x] = 0;
  
}

signed main(){
  
  cin>>n>>m;
  
  for(int i=0;i<m;i++){
    int a, b;
    cin>>a>>b;
    G[a-1].push_back(b-1);
    rG[b-1].push_back(a-1);
  }
  
  for(int i=0;i<n;i++){
    start = i;
    memset( used, 0, sizeof(used) );
    dfs(i,-1);
    if( f ) break;
  }
  
  if( !f ){
    cout<<-1<<endl;
    return 0;
  }
  
  int cnt = 0;
  
  for(int i=0;i<n;i++){
    if( line[i] ) cnt++;
  }
  
  cout<<cnt<<endl;
  
  for(int i=0;i<n;i++){
    if( line[i] ) cout<<i+1<<endl;
  }
  
  return 0;
}

```