#include<iostream>
#include<vector>
using namespace std;
int N,M;
vector<int>G[1000];
vector<int>id;
int used[1000];
bool dfs(int u,int d)
{
	used[u]=d;
	for(int v:G[u])
	{
		if(used[v]||dfs(v,d+1))return true;
	}
	used[u]=0;
	return false;
}
int main()
{
	cin>>N>>M;
	for(int i=0;i<M;i++)
	{
		int u,v;cin>>u>>v;
		u--,v--;
		G[u].push_back(v);
	}
	for(int i=0;i<N;i++)id.push_back(i);
	bool flag=false;
	int ps;
	for(int i:id)
	{
		if(dfs(i,1))
		{
			flag=true;
			ps=i;
			break;
		}
	}
	if(!flag)
	{
		cout<<-1<<endl;
		return 0;
	}
	while(true)
	{
		vector<int>nxtid;
		for(int i:id)
		{
			if(used[i])
			{
				nxtid.push_back(i);
				vector<int>nxt;
				for(int v:G[i])if(used[v])nxt.push_back(v);
				G[i]=nxt;
			}
		}
		id=nxtid;
		int s=-1,t;
		for(int i:id)
		{
			for(int v:G[i])
			{
				if(used[i]+1!=used[v]&&v!=ps)
				{
					s=i;
					t=v;
					break;
				}
			}
			if(s>=0)break;
		}
		if(s<0)
		{
			cout<<id.size()<<endl;
			for(int i:id)cout<<i+1<<endl;
			return 0;
		}
		for(int i:id)used[i]=0;
		used[s]=1;
		dfs(t,2);
		ps=s;
	}
}
### 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<vector>
using namespace std;
int N,M;
vector<int>G[1000];
vector<int>id;
int used[1000];
bool dfs(int u,int d)
{
	used[u]=d;
	for(int v:G[u])
	{
		if(used[v]||dfs(v,d+1))return true;
	}
	used[u]=0;
	return false;
}
int main()
{
	cin>>N>>M;
	for(int i=0;i<M;i++)
	{
		int u,v;cin>>u>>v;
		u--,v--;
		G[u].push_back(v);
	}
	for(int i=0;i<N;i++)id.push_back(i);
	bool flag=false;
	int ps;
	for(int i:id)
	{
		if(dfs(i,1))
		{
			flag=true;
			ps=i;
			break;
		}
	}
	if(!flag)
	{
		cout<<-1<<endl;
		return 0;
	}
	while(true)
	{
		vector<int>nxtid;
		for(int i:id)
		{
			if(used[i])
			{
				nxtid.push_back(i);
				vector<int>nxt;
				for(int v:G[i])if(used[v])nxt.push_back(v);
				G[i]=nxt;
			}
		}
		id=nxtid;
		int s=-1,t;
		for(int i:id)
		{
			for(int v:G[i])
			{
				if(used[i]+1!=used[v]&&v!=ps)
				{
					s=i;
					t=v;
					break;
				}
			}
			if(s>=0)break;
		}
		if(s<0)
		{
			cout<<id.size()<<endl;
			for(int i:id)cout<<i+1<<endl;
			return 0;
		}
		for(int i:id)used[i]=0;
		used[s]=1;
		dfs(t,2);
		ps=s;
	}
}
```