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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using P=pair<int,int>;
#define rep(i,m,n) for(int i=m;i<n;i++)

int main() {
  int n,m;
  cin>>n>>m;
  vector<int> to[n],from[n];
  rep(i,0,m){
    int a,b;
    cin>>a>>b;
    to[a-1].push_back(b-1);
    from[b-1].push_back(a-1);
  }
  vector<int> ans(n+1);
  rep(i,0,n){
    int dist[n];
    rep(j,0,n) dist[j]=114514;
    dist[i]=0;
    queue<int> q;
    q.push(i);
    int par[n];
    rep(j,0,n) par[j]=-1;
    while(q.size()){
      int r=q.front();
      q.pop();
      for(int x:to[r]){
        if(dist[x]==114514){
          dist[x]=dist[r]+1;
          q.push(x);
          par[x]=r;
        }
      }
    }
    P p=P(114514,-1);
    for(int x:from[i]){
      if(dist[x]<p.first){
        p=P(dist[x],x);
      }
    }
    if(p.first<ans.size()){
      vector<int> v;
      int cur=p.second;
      while(cur!=i){
        v.push_back(cur);
        cur=par[cur];
      }
      v.push_back(i);
      ans=v;
    }
  }
  if(ans.size()==n+1) cout<<-1<<endl;
  else{
    cout<<ans.size()<<endl;
    for(int x:ans) cout<<x+1<<endl;
  }
}