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;
#define rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)a;i<(int)(b);++i)


void solve()
{
  int N, M;
  cin >> N >> M;
  vector<int> G[N];
  rep(i,M){
    int a, b;
    cin >> a >> b;
    a--; b--;
    G[a].push_back(b);
  }
  vector<int> ans;
  rep(i,N){
    vector<bool> used(N);
    queue<int> Q;
    queue<vector<int>> P;
    used[i] = true;
    Q.push(i);
    vector<int> tmp;
    tmp.push_back(i);
    P.push(tmp);
    while(Q.size()){
      int u = Q.front();
      Q.pop();
      auto p = P.front();
      P.pop();

      for(auto v : G[u]){
	if(used[v]){
	  //cout << v <<  " " << p << endl;
	  if(v!=p[0])continue;
	  if(ans.size()==0 || p.size()<ans.size()){
	    ans=p;	    
	  }
	}
	else{
	  used[v]=true;
	  Q.push(v);
	  p.push_back(v);
	  P.push(p);
	  p.pop_back();
	}
      }
    }
  }
  if(ans.size()==0)cout << -1 << endl;
  else{
    cout << ans.size() << endl;
    for(auto v : ans){
      cout << (v+1) << endl;
    }
  }
}
int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(11);
  solve();
  return 0; 
}