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:
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
const int N = 1000, INF = 1 << 30;
vector<int> g[N];
signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
  }
  int siz = 1 << 30;
  vector<int> ans;
  for (int st = 0; st < n; st++) {
    queue<int> que;
    que.push(st);
    vector<int> dist(n, INF), prv(n, -1);
    dist[st] = 0;
    while (que.size()) {
      int cur = que.front();
      que.pop();
      for (int nxt : g[cur]) {
        if (nxt == st) {
          if (siz < dist[cur]) continue;
          siz = dist[cur];
          int j = cur;
          ans.clear();
          ans.push_back(st);
          while (prv[j] != -1) {
            ans.push_back(j);
            j = prv[j];
          }
        } else if (dist[nxt] < INF) {
          continue;
        } else {
          que.push(nxt);
          dist[nxt] = dist[cur] + 1;
          prv[nxt] = cur;
        }
      }
    }
  }
  if (!ans.empty()) {
    cout << ans.size() << '\n';
    for (int i : ans) {
      cout << i + 1 << '\n';
    } 
  } else {
    cout << -1 << '\n';
  }
  return 0;
}