#include<bits/stdc++.h>
using namespace std;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}


template<typename T> void drop(const T &x){cout<<x<<endl;exit(0);}

//INSERT ABOVE HERE
signed main(){
  Int n,m;
  cin>>n>>m;
  vector< vector<Int> > G(n);
  for(Int i=0;i<m;i++){
    Int a,b;
    cin>>a>>b;
    a--;b--;
    G[a].emplace_back(b);
  }

  vector<Int> ans;
  for(Int s=0;s<n;s++){
    vector<Int> dp(n,-1),pr(n,-1);
    queue<Int> que;
    dp[s]=0;
    que.emplace(s);
    Int len=n+1,t=-1;
    while(!que.empty()){
      Int v=que.front();que.pop();
      for(Int u:G[v]){
        if(u==s&&len>dp[v]+1) len=dp[v],t=v;
        if(~dp[u]) continue;
        dp[u]=dp[v]+1;
        pr[u]=v;
        que.emplace(u);
      }
    }
    if(t<0) continue;

    if(ans.empty()||len<(Int)ans.size()){
      ans.clear();
      Int v=t;
      while(v!=s) ans.emplace_back(v),v=pr[v];
      ans.emplace_back(v);
    }
  }
  if(ans.empty()) drop(-1);
  cout<<ans.size()<<endl;
  for(Int v:ans) cout<<v+1<<endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}


template<typename T> void drop(const T &x){cout<<x<<endl;exit(0);}

//INSERT ABOVE HERE
signed main(){
  Int n,m;
  cin>>n>>m;
  vector< vector<Int> > G(n);
  for(Int i=0;i<m;i++){
    Int a,b;
    cin>>a>>b;
    a--;b--;
    G[a].emplace_back(b);
  }

  vector<Int> ans;
  for(Int s=0;s<n;s++){
    vector<Int> dp(n,-1),pr(n,-1);
    queue<Int> que;
    dp[s]=0;
    que.emplace(s);
    Int len=n+1,t=-1;
    while(!que.empty()){
      Int v=que.front();que.pop();
      for(Int u:G[v]){
        if(u==s&&len>dp[v]+1) len=dp[v],t=v;
        if(~dp[u]) continue;
        dp[u]=dp[v]+1;
        pr[u]=v;
        que.emplace(u);
      }
    }
    if(t<0) continue;

    if(ans.empty()||len<(Int)ans.size()){
      ans.clear();
      Int v=t;
      while(v!=s) ans.emplace_back(v),v=pr[v];
      ans.emplace_back(v);
    }
  }
  if(ans.empty()) drop(-1);
  cout<<ans.size()<<endl;
  for(Int v:ans) cout<<v+1<<endl;
  return 0;
}

```