#include <bits/stdc++.h>
using namespace std;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n, m; cin >> n >> m;
  VV<> g(n);
  while (m--) {
    int u, v; cin >> u >> v, --u, --v;
    g[u].push_back(v);
  }

  int mn = 1e9;
  V<> res;
  for (int r = 0; r < n; ++r) {
    queue<int> q;
    V<> d(n, -1), prv(n, -1);
    d[r] = 0;
    q.push(r);
    [&] {
      while (!q.empty()) {
        int v = q.front(); q.pop();
        for (int u : g[v]) if (d[u] == -1) {
          d[u] = d[v] + 1;
          prv[u] = v;
          q.push(u);
        } else if (u == r and d[v] + 1 < mn) {
          mn = d[v] + 1;
          V<>().swap(res);
          int x = v;
          while (true) {
            res.push_back(x);
            if (x == r) return;
            x = prv[x];
          }
        }
      }
    }();
  }

  if (res.empty()) {
    cout << -1 << '\n';
  } else {
    cout << res.size() << '\n';
    for (int v : res) {
      cout << v + 1 << '\n';
    }
  }
}
### Prompt

Develop a solution in CPP 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 lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n, m; cin >> n >> m;
  VV<> g(n);
  while (m--) {
    int u, v; cin >> u >> v, --u, --v;
    g[u].push_back(v);
  }

  int mn = 1e9;
  V<> res;
  for (int r = 0; r < n; ++r) {
    queue<int> q;
    V<> d(n, -1), prv(n, -1);
    d[r] = 0;
    q.push(r);
    [&] {
      while (!q.empty()) {
        int v = q.front(); q.pop();
        for (int u : g[v]) if (d[u] == -1) {
          d[u] = d[v] + 1;
          prv[u] = v;
          q.push(u);
        } else if (u == r and d[v] + 1 < mn) {
          mn = d[v] + 1;
          V<>().swap(res);
          int x = v;
          while (true) {
            res.push_back(x);
            if (x == r) return;
            x = prv[x];
          }
        }
      }
    }();
  }

  if (res.empty()) {
    cout << -1 << '\n';
  } else {
    cout << res.size() << '\n';
    for (int v : res) {
      cout << v + 1 << '\n';
    }
  }
}
```