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;

typedef long long ll;
typedef pair<int, int> pii;

#define fi first
#define se second
#define mp make_pair
#define fastIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

const int N = (int)1005;
vector<int> T[N];
int ai[2 * N], bi[2 * N];
int pr[N];


int main(){
  fastIO;
  int n, m;
  cin >> n >> m;
  for(int i = 0 ; i < m ; i ++ ){
    cin >> ai[i] >> bi[i];
    T[ai[i]].push_back(bi[i]);
  }
  int nd;
  vector<int> ret;
  for(int i = 1; i <= n; i ++ ){
    for(int j = 1; j <= n; j ++ )
      pr[j] = -1;
    queue<int> qq;
    qq.push(i);
    while(!qq.empty()){
      nd = qq.front();
      qq.pop();
      for(auto x : T[nd]){
        if(pr[x] == -1){
          pr[x] = nd;
          qq.push(x);
        }
      }
    }
    if(pr[i] != -1){
      vector<int> res;
      nd = pr[i];
      while(nd != i){
        res.push_back(nd);
        nd = pr[nd];
      }
      res.push_back(nd);
      reverse(res.begin(), res.end());
      if(ret.empty() || res.size() < ret.size())
        ret = res;
    }
  }
  if(ret.empty()){
    cout << -1;
    return 0;
  }
  cout << ret.size() << "\n";
  for(auto x : ret)
    cout << x << " ";
  return 0;
}