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;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=5e15;

const ll N=1e3+10;
ll n,m;
vector<set<ll>> G(N),GG(N);

int main() {
  cin >> n >> m;
  for(ll i=0;i<m;i++) {
    ll a,b;
    cin >> a >> b;
    a--,b--;
    G[a].insert(b);
    GG[b].insert(a);
  }
  ll ans=inf;
  vec ansv(n);
  for(ll i=0;i<n;i++) {
    priority_queue<pll> q;
    ll id=-1;
    vec d(n,-1);
    q.push({0,i});
    while(q.size()) {
      ll f=q.top().second;
      ll td=q.top().first;
      q.pop();
      if(d[f]>-1) continue;
      d[f]=td;
      for(ll t:G[f]) {
        ll nd=td+1;
        if(t==i) {
          id=f;
          break;
        }
        q.push({nd,t});
      }
      if(id>-1) break;
    }
    if(id>-1) {
      if(ans>d[id]+1&&d[id]) {
        ans=d[id]+1;
        ll j=0;
        while(1) {
          ansv[j]=id+1;
          j++;
          if(id==i) break;
          for(ll t:GG[id]) {
            if(d[t]+1==d[id]) {
              id=t;
              break;
            }
          }
        }
      }
    }
  }
  if(ans==inf) ans=-1;
  cout << ans << endl;
  for(ll i=0;i<ans;i++) {
    cout << ansv[ans-i-1] << endl;
  }
}