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 int ll;

const int N = 1e3 + 5;

vector < vector < int > > adj (N);

int main () {
  ios_base::sync_with_stdio(false);
  cin.tie(0);cout.tie(0);

  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back (v);
  }

  int len = INT_MAX;
  vector < int > ans;
  for (int s = 1; s <= n; s++) {
    queue < int > q;
    q.push (s);
    vector < bool > used (n + 1, false);
    vector < int > p (n + 1);
    used[s] = true;
    while (!q.empty()) {
      int u = q.front(); q.pop();
      for (auto to : adj[u]) {
        if (!used[to]) {
          p[to] = u;
          used[to] = true;
          q.push (to);
        } else if (to == s) {
          vector < int > path;
          for (int i = u; i != s; i = p[i]) path.push_back (i);
          path.push_back (s);
          if ((int)path.size() < len) {
            len = (int)path.size();
            ans = path;
          }
        }
      }
    }
  }

  if (len == INT_MAX) {
    cout << -1 << '\n';
  } else {
    cout << len << '\n';
    for (auto i : ans) cout << i << '\n';
  }
}