#include<iostream>
#include<queue>
#include<memory>
#include<map>
#include<set>
#include<unordered_map>
#include<array>
#include<vector>
#include<algorithm>
#include<list>

using namespace std;
using vec = vector<int>;
using mat = vector<vec>;
using lst = list<int>;

int main()
{
	int N, M;
	cin >> N >> M;

	mat edge(N);

	for (int i = 0; i < M; ++i)
	{
		int A, B;
		cin >> A >> B;
		--A; --B;
		edge[A].emplace_back(B);
	}


	for (int s = 0; s < N; ++s)
	{
		lst Q;
		vec used(N, -1);
		
		Q.emplace_back(s);
		used[s] = s;

		bool flag = true;

		while(!Q.empty()&&flag)
		{
			int v = Q.front(); Q.pop_front();
			
			for (int u : edge[v])
			{
				if (used[u] == -1)
				{
					Q.emplace_back(u);
					used[u] = v;
				}
				else if (u == s)
				{
					set<int> st;
					lst ans={s};
					
					while (v != s)
					{
						if (st.count(v)) { flag = false; break; }

						ans.emplace_back(v);
						for (auto w : edge[v])
						{
							if (w != used[v]) { st.insert(w); }
						}
						v = used[v];
					}

					if (flag)
					{
						cout << ans.size() << endl;
						for (int a : ans) { cout << a+1 << endl; }
						return 0;
					}

					break;
				}
			}
		}
	}

	cout << -1 << endl;

	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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<memory>
#include<map>
#include<set>
#include<unordered_map>
#include<array>
#include<vector>
#include<algorithm>
#include<list>

using namespace std;
using vec = vector<int>;
using mat = vector<vec>;
using lst = list<int>;

int main()
{
	int N, M;
	cin >> N >> M;

	mat edge(N);

	for (int i = 0; i < M; ++i)
	{
		int A, B;
		cin >> A >> B;
		--A; --B;
		edge[A].emplace_back(B);
	}


	for (int s = 0; s < N; ++s)
	{
		lst Q;
		vec used(N, -1);
		
		Q.emplace_back(s);
		used[s] = s;

		bool flag = true;

		while(!Q.empty()&&flag)
		{
			int v = Q.front(); Q.pop_front();
			
			for (int u : edge[v])
			{
				if (used[u] == -1)
				{
					Q.emplace_back(u);
					used[u] = v;
				}
				else if (u == s)
				{
					set<int> st;
					lst ans={s};
					
					while (v != s)
					{
						if (st.count(v)) { flag = false; break; }

						ans.emplace_back(v);
						for (auto w : edge[v])
						{
							if (w != used[v]) { st.insert(w); }
						}
						v = used[v];
					}

					if (flag)
					{
						cout << ans.size() << endl;
						for (int a : ans) { cout << a+1 << endl; }
						return 0;
					}

					break;
				}
			}
		}
	}

	cout << -1 << endl;

	return 0;
}
```