/*
luogu AT5256
*/
#include <bits/stdc++.h>
using namespace std;

#define int long long

int n,m;
vector<int> ans;
int ee,h[100005],nex[100005],to[100005];
int vis[100005],in[100005],top,sk[100005];
int g[1005][1005];

void addedge(int x,int y)
{
	nex[++ee] = h[x];
	to[ee] = y;
	h[x] = ee;
}

void dfs(int x)
{
	vis[x] = true;
	sk[++top] = x;
	in[x] = true;
	for(int i=h[x];i;i=nex[i])
	{
		if(in[to[i]])
		{
			do{
				ans.push_back(sk[top]);
			}while(sk[top--]!=to[i]);
			cout<<ans.size()<<endl;
			for(int i=0;i<ans.size();i++)
				cout<<ans[i]<<endl;
			exit(0);
		}
		else if(!vis[to[i]])
			dfs(to[i]);
	}
	in[x] = false;
	top--;
}

signed main()
{
	cin>>n>>m;
	for(int i=1,x,y;i<=m&&cin>>x>>y;i++)
		addedge(x,y), g[x][y] = 1;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if(g[i][j] && g[j][i])
				return cout<<2<<endl<<i<<endl<<j<<endl, 0;
	for(int i=1;i<=n;i++)
		if(!vis[i])
			dfs(i);
	puts("-1");
	
	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
/*
luogu AT5256
*/
#include <bits/stdc++.h>
using namespace std;

#define int long long

int n,m;
vector<int> ans;
int ee,h[100005],nex[100005],to[100005];
int vis[100005],in[100005],top,sk[100005];
int g[1005][1005];

void addedge(int x,int y)
{
	nex[++ee] = h[x];
	to[ee] = y;
	h[x] = ee;
}

void dfs(int x)
{
	vis[x] = true;
	sk[++top] = x;
	in[x] = true;
	for(int i=h[x];i;i=nex[i])
	{
		if(in[to[i]])
		{
			do{
				ans.push_back(sk[top]);
			}while(sk[top--]!=to[i]);
			cout<<ans.size()<<endl;
			for(int i=0;i<ans.size();i++)
				cout<<ans[i]<<endl;
			exit(0);
		}
		else if(!vis[to[i]])
			dfs(to[i]);
	}
	in[x] = false;
	top--;
}

signed main()
{
	cin>>n>>m;
	for(int i=1,x,y;i<=m&&cin>>x>>y;i++)
		addedge(x,y), g[x][y] = 1;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if(g[i][j] && g[j][i])
				return cout<<2<<endl<<i<<endl<<j<<endl, 0;
	for(int i=1;i<=n;i++)
		if(!vis[i])
			dfs(i);
	puts("-1");
	
	return 0;
}
```