#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<map>
using namespace std;

int from[100010],to[100010];
int n,m;
vector<int> V[100010];
int vis[100010],F[100010];
int ans;
map<int,map<int,int>> p;

int bfs(int st)
{
	queue<int> Q;
	memset(vis,-1,sizeof(vis));
	memset(F,0,sizeof(F));
	Q.push(st);
	vis[st]=1;
	F[st]=0;
	while(!Q.empty())
	{
		int cur=Q.front();
		Q.pop();
		for(auto k:V[cur])
		{
			int u=to[k];
			if(u==st) 
				return ans=cur;
			if(vis[u]!=-1) 
				continue;
			vis[u]=vis[cur]+1;
			F[u]=cur;
			Q.push(u);
		}
	}
	return ans=-1;
}

int main()
{
	cin>>n>>m;
	for(int i=1; i<=m; i++)
	{
		cin>>from[i]>>to[i];
		V[from[i]].push_back(i);
	}
	int len=12345678;
	int A=0;
	for(int i=1; i<=n; i++)
	{
		int tmp=bfs(i);
		if(tmp==-1) 
			continue;
		if(vis[tmp]<len)
		{
			A=i;
			len=vis[tmp];
		}
	}
	if(!A)
	{
		cout<<-1<<endl;
		return 0;
	}
	bfs(A);
	cout<<len<<endl;
	for(int i=1; i<=len; i++)
	{
		cout<<ans<<endl;
		ans=F[ans];
	}
	return 0;
}

### Prompt

Generate 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<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<map>
using namespace std;

int from[100010],to[100010];
int n,m;
vector<int> V[100010];
int vis[100010],F[100010];
int ans;
map<int,map<int,int>> p;

int bfs(int st)
{
	queue<int> Q;
	memset(vis,-1,sizeof(vis));
	memset(F,0,sizeof(F));
	Q.push(st);
	vis[st]=1;
	F[st]=0;
	while(!Q.empty())
	{
		int cur=Q.front();
		Q.pop();
		for(auto k:V[cur])
		{
			int u=to[k];
			if(u==st) 
				return ans=cur;
			if(vis[u]!=-1) 
				continue;
			vis[u]=vis[cur]+1;
			F[u]=cur;
			Q.push(u);
		}
	}
	return ans=-1;
}

int main()
{
	cin>>n>>m;
	for(int i=1; i<=m; i++)
	{
		cin>>from[i]>>to[i];
		V[from[i]].push_back(i);
	}
	int len=12345678;
	int A=0;
	for(int i=1; i<=n; i++)
	{
		int tmp=bfs(i);
		if(tmp==-1) 
			continue;
		if(vis[tmp]<len)
		{
			A=i;
			len=vis[tmp];
		}
	}
	if(!A)
	{
		cout<<-1<<endl;
		return 0;
	}
	bfs(A);
	cout<<len<<endl;
	for(int i=1; i<=len; i++)
	{
		cout<<ans<<endl;
		ans=F[ans];
	}
	return 0;
}

```