#include <bits/stdc++.h>
using namespace std;
int dist[1005],par[1005];
vector<int> v[1005];
pair<int,pair<int,int> > ans(1e9,{0,0});
void bfs(int i)
{
	memset(dist,0,sizeof(dist));
	dist[i]=1;
	par[i]=0;
	queue<int> q;
	q.push(i);
	while (!q.empty())
	{
		int node=q.front();
		q.pop();
		for (int u:v[node])
		{
			if (u==i)
			ans=min(ans,make_pair(dist[node],make_pair(node,i)));
			if (!dist[u])
			{
				q.push(u);
				dist[u]=dist[node]+1;
				par[u]=node;
			}
		}
	}
}
void go(int node)
{
	if (par[node])
	go(par[node]);
	printf("%d\n",node);
}
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	while (m--)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		v[a].push_back(b);
	}
	for (int i=1;i<=n;i++)
	bfs(i);
	if (ans.first==1e9)
	{
		printf("-1");
		return 0;
	}
	bfs(ans.second.second);
	printf("%d\n",ans.first);
	go(ans.second.first);
}
### Prompt

Your task is to create 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;
int dist[1005],par[1005];
vector<int> v[1005];
pair<int,pair<int,int> > ans(1e9,{0,0});
void bfs(int i)
{
	memset(dist,0,sizeof(dist));
	dist[i]=1;
	par[i]=0;
	queue<int> q;
	q.push(i);
	while (!q.empty())
	{
		int node=q.front();
		q.pop();
		for (int u:v[node])
		{
			if (u==i)
			ans=min(ans,make_pair(dist[node],make_pair(node,i)));
			if (!dist[u])
			{
				q.push(u);
				dist[u]=dist[node]+1;
				par[u]=node;
			}
		}
	}
}
void go(int node)
{
	if (par[node])
	go(par[node]);
	printf("%d\n",node);
}
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	while (m--)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		v[a].push_back(b);
	}
	for (int i=1;i<=n;i++)
	bfs(i);
	if (ans.first==1e9)
	{
		printf("-1");
		return 0;
	}
	bfs(ans.second.second);
	printf("%d\n",ans.first);
	go(ans.second.first);
}
```