#include <bits/stdc++.h>
 
using namespace std;
 
const int DIM = 1e5 + 7;
 
bitset <DIM> vis;
 
vector <int> st;
vector <int> v[DIM];
vector <int> res;
 
void dfs(int nod)
{
	vis[nod] = true;
	st.push_back(nod);
	
	for(auto i : v[nod])
		if(i == st[0] && nod != st[0] && (res.size() == 0 || res.size() > st.size()))
		{
			res = st;
		}
		else
			if(vis[i] == 0)
			{
				vis[i] = true;
				dfs(i);
			}
	
	st.pop_back();
}
 
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	int n, m;
	cin >> n >> m;
	
	for(int i = 1; i <= m; i++)
	{
		int x, y;
		cin >> x >> y;
		
		v[x].push_back(y);
	}
	
	for(int i = 1; i <= n; i++)
	{
		vis.reset();
		dfs(i);
	}
	if(res.size() == 0)
		cout << -1;
	else
	{
		cout << res.size() << '\n';
		
		for(auto i : res)
			cout << i << '\n';
	}
}
### Prompt

Please formulate 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;
 
const int DIM = 1e5 + 7;
 
bitset <DIM> vis;
 
vector <int> st;
vector <int> v[DIM];
vector <int> res;
 
void dfs(int nod)
{
	vis[nod] = true;
	st.push_back(nod);
	
	for(auto i : v[nod])
		if(i == st[0] && nod != st[0] && (res.size() == 0 || res.size() > st.size()))
		{
			res = st;
		}
		else
			if(vis[i] == 0)
			{
				vis[i] = true;
				dfs(i);
			}
	
	st.pop_back();
}
 
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	int n, m;
	cin >> n >> m;
	
	for(int i = 1; i <= m; i++)
	{
		int x, y;
		cin >> x >> y;
		
		v[x].push_back(y);
	}
	
	for(int i = 1; i <= n; i++)
	{
		vis.reset();
		dfs(i);
	}
	if(res.size() == 0)
		cout << -1;
	else
	{
		cout << res.size() << '\n';
		
		for(auto i : res)
			cout << i << '\n';
	}
}
```