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) for(int i = 0; i < (int)(n); i++)

int main(){
  int n, m;
  cin >> n >> m;
  vector<vector<int> > A(n);
  REP(i, m){
    int a, b;
    cin >> a >> b;
    --a, --b;
    A[a].push_back(b);
  }
  vector<vector<int> > B;
  vector<vector<int> > T;
  REP(i, n){
    queue<pair<int, int> > q;
    q.push(make_pair(i, 0));
    vector<int> v(n, -1);
    vector<int> t(n);
    while(!q.empty()){
      auto x = q.front();
      q.pop();
      for(auto y: A[x.first]){
        if(v[y] >= 0) continue;
        t[y] = x.first;
        v[y] = x.second + 1;
        q.push(make_pair(y, x.second + 1));
      }
    }
    B.push_back(v);
    T.push_back(t);
  }
  int r = 10000000;
  int mi, mj;
  REP(i, n){
    if(B[i][i] > 0){
      if(B[i][i] < r){
        mi = i;
        r = B[i][i];
      }
    }
  }
  if(r > n){
    cout << -1 << endl;
    return 0;
  }
  cout << r << endl;
  vector<int> ans(r);
  auto t = T[mi];
  int y = mi;
  REP(i, r){
    ans[r - i - 1] = y;
    y = t[y];
  }
  REP(i, r){
    cout << ans[i] + 1 << endl;
  }
  return 0;
}