#include <iostream>
#include <vector>
#include <queue>

using namespace std;

vector<int> g[1005];

int main() {
  int n,m;
  cin >> n >> m;
  for(int i = 0; i < m; i++) {
    int a,b;
    cin >> a >> b;
    --a;--b;
    g[a].push_back(b);
  }

  vector<int> ans(n+1);
  int M = ans.size();
  
  for(int s = 0; s < n; s++) {
    vector<int> dist(n,-1);
    vector<int> prev(n,-1);
    queue<int> q;
    dist[s] = 0;
    q.push(s);
    while(!q.empty()) {
      int v = q.front();
      q.pop();
      for(int i = 0; i < g[v].size(); i++) {
	int u = g[v][i];
	if(dist[u] != -1) continue;
	dist[u] = dist[v] + 1;
	prev[u] = v;
	q.push(u);
      }
    }

    for(int t = 0; t < n; t++) {
      if(t == s || dist[t] == -1) continue;
      for(int i = 0; i < g[t].size(); i++) {
	int u = g[t][i];
	if(u == s) {
	  vector<int> tmp;
	  tmp.push_back(s);
	  int v = t;
	  while(v != s) {
	    tmp.push_back(v);
	    v = prev[v];
	  }
	  if(M > tmp.size()) {
	    M = tmp.size();
	    ans = tmp;
	  }
	}
      }
    }
  }

  if(M == n + 1) {
    cout << -1 << endl;
    return 0;
  }
  else {
    cout << M << endl;
    for(int i = 0; i < ans.size(); i++) {
      cout << ans[i] + 1 << endl;
    }
  }
}

### Prompt

Please create a solution in CPP 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 <iostream>
#include <vector>
#include <queue>

using namespace std;

vector<int> g[1005];

int main() {
  int n,m;
  cin >> n >> m;
  for(int i = 0; i < m; i++) {
    int a,b;
    cin >> a >> b;
    --a;--b;
    g[a].push_back(b);
  }

  vector<int> ans(n+1);
  int M = ans.size();
  
  for(int s = 0; s < n; s++) {
    vector<int> dist(n,-1);
    vector<int> prev(n,-1);
    queue<int> q;
    dist[s] = 0;
    q.push(s);
    while(!q.empty()) {
      int v = q.front();
      q.pop();
      for(int i = 0; i < g[v].size(); i++) {
	int u = g[v][i];
	if(dist[u] != -1) continue;
	dist[u] = dist[v] + 1;
	prev[u] = v;
	q.push(u);
      }
    }

    for(int t = 0; t < n; t++) {
      if(t == s || dist[t] == -1) continue;
      for(int i = 0; i < g[t].size(); i++) {
	int u = g[t][i];
	if(u == s) {
	  vector<int> tmp;
	  tmp.push_back(s);
	  int v = t;
	  while(v != s) {
	    tmp.push_back(v);
	    v = prev[v];
	  }
	  if(M > tmp.size()) {
	    M = tmp.size();
	    ans = tmp;
	  }
	}
      }
    }
  }

  if(M == n + 1) {
    cout << -1 << endl;
    return 0;
  }
  else {
    cout << M << endl;
    for(int i = 0; i < ans.size(); i++) {
      cout << ans[i] + 1 << endl;
    }
  }
}

```