#include<bits/stdc++.h>
using namespace std;
const int N=1011;
int used[N],br[N],l[N];
vector<int>v[N],ans;
bool dfs(int x,int z)
{
	used[x]=1;
	bool l,l2=1;
	for(auto y:v[x])
	{
		br[y]++;
		if(used[y]==1 && br[y]>1)l2=0;
	}
	if(l2)
		for(auto y:v[x])
		{
			if(y==z && br[z]==1)
			{
				ans.push_back(x);
				return 1;
			}
			else if(br[y]==1 && used[y]==0)
			{
				l=dfs(y,z);
				if(l)
				{
					ans.push_back(x);
					return 1;
				}
			}
		}
	for(auto y:v[x])
		br[y]--;
	used[x]=2;
	return 0;
}
int main()
{
     ios_base::sync_with_stdio(0);
     cin.tie(0);
     int n,m,i,a,b;
     cin>>n>>m;
	 for(i=0;i<m;i++)
	 {
		 cin>>a>>b;
		 v[a].push_back(b);
	 }
	 bool l;
	 for(i=1;i<=n;i++)
	 {
		 memset(used,0,sizeof(used));
		 l=dfs(i,i);
		 if(l)break;
	 }
	 if(ans.size()==0)
	 {
		 cout<<"-1\n";
		 return 0;
	 }
	 cout<<ans.size()<<'\n';
	 for(auto x:ans)
		 cout<<x<<'\n';
}

### Prompt

Your challenge is to write 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 N=1011;
int used[N],br[N],l[N];
vector<int>v[N],ans;
bool dfs(int x,int z)
{
	used[x]=1;
	bool l,l2=1;
	for(auto y:v[x])
	{
		br[y]++;
		if(used[y]==1 && br[y]>1)l2=0;
	}
	if(l2)
		for(auto y:v[x])
		{
			if(y==z && br[z]==1)
			{
				ans.push_back(x);
				return 1;
			}
			else if(br[y]==1 && used[y]==0)
			{
				l=dfs(y,z);
				if(l)
				{
					ans.push_back(x);
					return 1;
				}
			}
		}
	for(auto y:v[x])
		br[y]--;
	used[x]=2;
	return 0;
}
int main()
{
     ios_base::sync_with_stdio(0);
     cin.tie(0);
     int n,m,i,a,b;
     cin>>n>>m;
	 for(i=0;i<m;i++)
	 {
		 cin>>a>>b;
		 v[a].push_back(b);
	 }
	 bool l;
	 for(i=1;i<=n;i++)
	 {
		 memset(used,0,sizeof(used));
		 l=dfs(i,i);
		 if(l)break;
	 }
	 if(ans.size()==0)
	 {
		 cout<<"-1\n";
		 return 0;
	 }
	 cout<<ans.size()<<'\n';
	 for(auto x:ans)
		 cout<<x<<'\n';
}

```