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>
#define rep(i,n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

int n;
vector<vector<int>> g;

vector<int> bfs(int s)
{
  vector<int> ret, par(n, -1);
  queue<int> q;
  q.push(s);
  while (!q.empty())
  {
    int crt = q.front();
    q.pop();
    for (int nxt : g[crt])
    {
      if (par[nxt] != -1)
        continue;
      par[nxt] = crt;
      q.push(nxt);
    }
  }
  if (par[s] == -1)
    return vector<int>(n + 1);
  ret.push_back(par[s] + 1);
  int p = par[s];
  while (p != s)
  {
    p = par[p];
    ret.push_back(p + 1);
  }
  return ret;
}

int main()
{
  int m;
  cin >> n >> m;
  g.resize(n);
  rep(i, m)
  {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
  }
  vector<int> ans(n + 1);
  rep(i, n)
  {
    vector<int> tmp = bfs(i);
    if (tmp.size() < ans.size())
      ans.swap(tmp);
  }
  if (ans.size() == n + 1)
  {
    puts("-1");
    return 0;
  }
  printf("%d\n", ans.size());
  for (int i : ans)
    printf("%d\n", i);
  return 0;
}