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;

const int INF = 1001001001;
int n;
vector<int> to[1005];

vector<int> bfs(int sv){
  vector<int> dist(n, INF), pre(n, -1);
  queue<int> q;
  dist[sv] = 0;
  q.push(sv);
  while (!q.empty()) {
    int v = q.front(); q.pop();
    for (int u : to[v]) {
      if (dist[u] != INF) continue;
      pre[u] = v;
      dist[u] = dist[v]+1;
      q.push(u);
    }
  }

  pair<int, int> best(INF, -1);
  rep(v, n) {
    if (v == sv) continue;
    for (int u : to[v]) {
      if (u == sv) {
        best = min(best, make_pair(dist[v],v));
      }
    }
  }
  if (best.first == INF) return vector<int>(n+1);
  int v = best.second;
  vector<int> res;
  while (v != -1){
    res.push_back(v);
    v = pre[v];
  }
  return res;
}

int main(){
  int m; cin >> n >> m;
  rep(i,m){
    int a,b; cin >> a >> b;
    --a; --b;
    to[a].push_back(b);
  }

  vector<int> ans(n+1);
  rep(s, n) {
    vector<int> now = bfs(s);
    if (now.size() < ans.size()) ans = now;
  }
  if (ans.size() == n+1) {
    cout << -1 << endl;
    return 0;
  }
  cout << ans.size() << endl;
  for (int v : ans){
    cout << v+1 << endl;
  }
  
  return 0;
}