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 ll;
#define rep(i, N) for (int i = 0; i < (int)N; ++i)
#define all(a) (a).begin(), (a).end()

int N, M;
const int INF = 1e9;
int S = 1e9;
vector<vector<int>> G;
stack<int> ans;

void search(int s) {
  vector<int> d(N, INF), prev(N, -1);
  queue<int> q;
  q.push(s);

  d[s] = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();

    for (int v : G[u]) {
      if (v == s) {
        if (S > d[u]) {
          S = d[u];
          ans = stack<int>();
          while (u != -1) {
            ans.push(u);
            u = prev[u];
          }
          return;
        }
      }
      if (d[v] > d[u] + 1) {
        prev[v] = u;
        d[v] = d[u] + 1;
        q.push(v);
      }
    }
  }
}

int main() {
  cin >> N >> M;
  G.resize(N);
  rep(i, M) {
    int A, B;
    cin >> A >> B;
    A--, B--;
    G[A].push_back(B);
  }

  rep(i, N) search(i);

  if (ans.size() == 0) {
    cout << -1 << endl;
    return 0;
  }

  cout << ans.size() << endl;
  while (!ans.empty()) {
    cout << (ans.top() + 1) << endl;
    ans.pop();
  }

  return 0;
}