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() {
  int n, m;
  cin >> n >> m;
  vector<vector<int>> v(n);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    v[a - 1].push_back(b - 1);
  }

  vector<int> d(n), in(n), s, r;
  function<void(int)> dfs = [&](int i) {
    d[i] = s.size();
    s.push_back(i);
    for (auto j : v[i]) in[j]++;

    priority_queue<pair<int, int>> q;
    for (auto j : v[i]) {
      if (d[j] != -1) q.push(make_pair(d[j], j));
    }
    if (!q.empty()) {
      r = vector<int>(s.begin() + q.top().first, s.end());
      return;
    }
    for (auto j : v[i]) {
      if (d[j] != -1 || in[j] > 1) continue;
      dfs(j);
      if (!r.empty()) return;
    }

    for (auto j : v[i]) in[j]--;
    s.pop_back();
    d[i] = -1;
  };
  for (int i = 0; i < n; i++) {
    fill(d.begin(), d.end(), -1);
    fill(in.begin(), in.end(), 0);
    dfs(i);
    if (!r.empty()) break;
  }

  cout << (r.empty() ? -1 : (int)r.size()) << endl;
  for (auto e : r) cout << e + 1 << endl;
}