#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long double ld;

const int N = 1010;

int n,m;
vector<int> adj[N];
int vis[N];

vector<int> ans;

void dfs(int u, int ra, vector<int>& sta) {
  if (vis[u]) {
    if (u == ra && (ans.empty() || ans.size() > sta.size())) {
      ans = sta;
    }
    return;
  }
  vis[u] = 1;
  sta.push_back(u);

  for (auto& v : adj[u]) {
    dfs(v, ra, sta);
  }

  sta.pop_back();
}

void check(int u) {
  memset(vis, 0, sizeof vis);
  vector<int> sta;
  dfs(u, u, sta);
}

int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m;
  for (int i=1; i<=m; ++i) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
  }

  for (int i=1; i<=n; ++i) {
    check(i);
  }

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

  cout << ans.size() << endl;
  for (auto& x : ans) {
    cout << x << "\n";
  }
  return 0; 
}

### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long double ld;

const int N = 1010;

int n,m;
vector<int> adj[N];
int vis[N];

vector<int> ans;

void dfs(int u, int ra, vector<int>& sta) {
  if (vis[u]) {
    if (u == ra && (ans.empty() || ans.size() > sta.size())) {
      ans = sta;
    }
    return;
  }
  vis[u] = 1;
  sta.push_back(u);

  for (auto& v : adj[u]) {
    dfs(v, ra, sta);
  }

  sta.pop_back();
}

void check(int u) {
  memset(vis, 0, sizeof vis);
  vector<int> sta;
  dfs(u, u, sta);
}

int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m;
  for (int i=1; i<=m; ++i) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
  }

  for (int i=1; i<=n; ++i) {
    check(i);
  }

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

  cout << ans.size() << endl;
  for (auto& x : ans) {
    cout << x << "\n";
  }
  return 0; 
}

```