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;

int main() {
  ios::sync_with_stdio(false); cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<vector<int> > adj(n + 1);
  vector<pair<int, int> > E;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    E.emplace_back(u, v);
    adj[u].push_back(v);
  }
  vector<int> ans;
  for (auto e: E) {
    vector<int> parent(n + 1), dist(n + 1, n + 1);
    queue<int> q;
    q.emplace(e.second);
    dist[e.second] = 0;
    parent[e.second] = -1;
    while (!q.empty()) {
      int u = q.front(); q.pop();
      for (auto v: adj[u]) {
        if (dist[u] + 1 < dist[v]) {
          dist[v] = dist[u] + 1;
          q.emplace(v);
          parent[v] = u;
        }
      }
    }
    if (dist[e.first] != n + 1) {
      if (ans.empty() || dist[e.first] + 1 < ans.size()) {
        ans.clear();
        int cur = e.first;
        while (cur != -1) {
          ans.push_back(cur);
          cur = parent[cur];
        }
      }
    }
  }
  if (ans.empty()) {
    cout << -1 << endl;
    return 0;
  }
  cout << ans.size() << endl;
  for (auto u: ans) cout << u << endl;
  return 0;
}