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 ll = int_fast64_t;
#define SELECTOR(_1, _2, _3, SELECT, ...) SELECT
#define rep(...) SELECTOR(__VA_ARGS__, _rep1, _rep0)(__VA_ARGS__)
#define _rep0(i, n) for (ll i = 0; i < n; ++i)
#define _rep1(i, k, n) for (ll i = k; i < n; ++i)

void solve(){
  // main coding
  ll n,m; cin>>n>>m;
  ll INF=1<<15;
  vector<vector<ll>> dist(n,vector<ll>(n,INF)); // adjacency matrix
  vector<vector<ll>> E(n); // adjacency list
  rep(_,m) {
    ll a,b; cin>>a>>b;
    --a; --b; // 0-indexed
    dist[a][b]=1;
    E[a].push_back(b);
  }
  // Warshall-Floyd
  rep(k,n) rep(i,n) rep(j,n) {
    dist[i][j]=min(dist[i][k]+dist[k][j],dist[i][j]);
  }
  // loop detection
  ll len=n+1,v;
  rep(i,n) {
    if(dist[i][i]<len) {
      v=i;
      len=dist[i][i];
    }
  }
  if(len==n+1) {
    cout<<-1<<"\n";
    return;
  }
  const ll vm=v;
  vector<ll> ans;
  while(len){
    ans.push_back(v);
    --len;
    rep(i,n) if(dist[v][i]==1 && dist[i][vm]==len){
      v=i;
      break;
    }
  }
  cout<<ans.size()<<"\n";
  for(ll v:ans) cout<<v+1<<"\n";
}

int main(){
  ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(10);
  solve();
  return 0;
}