
#include<bits/stdc++.h>
using namespace std;
using u64 = uint_fast64_t;
using i64 = int_fast64_t;
#define rep(i, n) for(u64 i = 0; i < n; ++i)
#define dump(x) cout << #x << " = " << x << endl
template<class T>
bool chmin(T &a, const T &b){
  if(b < a){
    a = b;
    return true;
  }
  return false;
}

int main(){
  u64 n, m;
  cin >> n >> m;
  vector<vector<u64>> g(n);
  vector<vector<u64>> dist(n, vector<u64>(n, u64(1) << 30));
  rep(i, m){
    u64 a, b;
    cin >> a >> b;
    --a;
    --b;
    g[a].push_back(b);
    dist[a][b] = 1;
  }
  rep(j, n){
    rep(i, n){
      rep(k, n){
        chmin(dist[i][k], dist[i][j] + dist[j][k]);
      }
    }
  }
  u64 minlen = n + 1, v = n;
  rep(i, n){
    if(chmin(minlen, dist[i][i])){
      v = i;
    }
  }
  if(v == n){
    cout << "-1" << endl;
    return 0;
  }
  const u64 vm = v;
  vector<u64> ans;
  ans.push_back(v);
  while(minlen){
    --minlen;
    rep(i, n){
      if(dist[v][i] == 1 && dist[i][vm] == minlen){
        ans.push_back(i);
        v = i;
        break;
      }
    }
  }
  cout << ans.size() << endl;
  for(const auto e : ans){
    cout << e + 1 << endl;
  }
  
  /*
  rep(i, n){
    rep(j, n){
      cout << dist[i][j] << ' ';
    }
    cout << endl;
  }
  */
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp

#include<bits/stdc++.h>
using namespace std;
using u64 = uint_fast64_t;
using i64 = int_fast64_t;
#define rep(i, n) for(u64 i = 0; i < n; ++i)
#define dump(x) cout << #x << " = " << x << endl
template<class T>
bool chmin(T &a, const T &b){
  if(b < a){
    a = b;
    return true;
  }
  return false;
}

int main(){
  u64 n, m;
  cin >> n >> m;
  vector<vector<u64>> g(n);
  vector<vector<u64>> dist(n, vector<u64>(n, u64(1) << 30));
  rep(i, m){
    u64 a, b;
    cin >> a >> b;
    --a;
    --b;
    g[a].push_back(b);
    dist[a][b] = 1;
  }
  rep(j, n){
    rep(i, n){
      rep(k, n){
        chmin(dist[i][k], dist[i][j] + dist[j][k]);
      }
    }
  }
  u64 minlen = n + 1, v = n;
  rep(i, n){
    if(chmin(minlen, dist[i][i])){
      v = i;
    }
  }
  if(v == n){
    cout << "-1" << endl;
    return 0;
  }
  const u64 vm = v;
  vector<u64> ans;
  ans.push_back(v);
  while(minlen){
    --minlen;
    rep(i, n){
      if(dist[v][i] == 1 && dist[i][vm] == minlen){
        ans.push_back(i);
        v = i;
        break;
      }
    }
  }
  cout << ans.size() << endl;
  for(const auto e : ans){
    cout << e + 1 << endl;
  }
  
  /*
  rep(i, n){
    rep(j, n){
      cout << dist[i][j] << ' ';
    }
    cout << endl;
  }
  */
}

```