#include <iostream>
#include <queue>
#include <stack>
#include <utility>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

const int INF = 1e7;
ll n, m, ans;
vector<ll> edge[1100];
ll pre[1100];
stack<ll> path;

int bfs(int s) {
  bool done[1100] = {};
  queue<P> q;
  
  q.push(P(s, 0));
  
  while (q.size()) {
    P p = q.front();
    q.pop();
    int u = p.first;
    int dist = p.second + 1;
    for (int v: edge[u]) {
      if (done[v]) continue;
      pre[v] = u;
      if (v == s) return dist;
      q.push(P(v, dist));
      done[v] = true;
    }
  }
  return INF;
}

int main() {
  cin >> n >> m;  
  for (int i = 0; i < m; i++) {
    ll a, b;
    cin >> a >> b;
    a--, b--;
    edge[a].push_back(b);
  }
  
  ll min_dist = INF, start;
  for (int i = 0; i < n; i++) {
    ll dist = bfs(i);
    if (dist < min_dist) {
      min_dist = dist;
      start = i;
    }
  }
  
  if (min_dist == INF) {
    cout << -1 << endl;
    return 0;
  }
  
  cout << min_dist << endl;
  
  bfs(start);
  path.push(pre[start]);
  while (path.top() != start) {
    path.push(pre[path.top()]);
  }
  while (path.size()) {
    cout << path.top() + 1 << endl;
    path.pop();
  }
  
  return 0;
}

### 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 <iostream>
#include <queue>
#include <stack>
#include <utility>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

const int INF = 1e7;
ll n, m, ans;
vector<ll> edge[1100];
ll pre[1100];
stack<ll> path;

int bfs(int s) {
  bool done[1100] = {};
  queue<P> q;
  
  q.push(P(s, 0));
  
  while (q.size()) {
    P p = q.front();
    q.pop();
    int u = p.first;
    int dist = p.second + 1;
    for (int v: edge[u]) {
      if (done[v]) continue;
      pre[v] = u;
      if (v == s) return dist;
      q.push(P(v, dist));
      done[v] = true;
    }
  }
  return INF;
}

int main() {
  cin >> n >> m;  
  for (int i = 0; i < m; i++) {
    ll a, b;
    cin >> a >> b;
    a--, b--;
    edge[a].push_back(b);
  }
  
  ll min_dist = INF, start;
  for (int i = 0; i < n; i++) {
    ll dist = bfs(i);
    if (dist < min_dist) {
      min_dist = dist;
      start = i;
    }
  }
  
  if (min_dist == INF) {
    cout << -1 << endl;
    return 0;
  }
  
  cout << min_dist << endl;
  
  bfs(start);
  path.push(pre[start]);
  while (path.top() != start) {
    path.push(pre[path.top()]);
  }
  while (path.size()) {
    cout << path.top() + 1 << endl;
    path.pop();
  }
  
  return 0;
}

```