#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#define ll long long
using namespace std;
bool visit[1005];
int st,stack[10050];
vector<int>side[1005];
int n,m,cnt,ans=1e9+1000,data[1005];
void dfs(int now){
//	cout<<now<<endl;
	stack[++cnt]=now;visit[now]=1;
	for(int i=0;i<side[now].size();i++){
		if(cnt>=ans){cnt--;return;}
		int v=side[now][i];
		if(v==st&&cnt<ans){
				ans=cnt;
				for(int j=cnt;j>=1;j--){data[j]=stack[j];if(stack[j]==v)break;}
				cnt--;
				return;
		}
		if(visit[v])continue;
		dfs(v);
	}
	cnt--;
}
int main(){
	ios::sync_with_stdio(0);
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		int u,v;cin>>u>>v;
		side[u].push_back(v);
	}
	for(int i=1;i<=n;i++){
		cnt=0;memset(visit,0,sizeof(visit));
		st=i;dfs(i);
	}
	if(ans<1e9){
		cout<<ans<<endl;
		for(int i=1;i<=ans;i++)cout<<data[i]<<endl;
		return 0;	
	}
	cout<<"-1";
}
### 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<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#define ll long long
using namespace std;
bool visit[1005];
int st,stack[10050];
vector<int>side[1005];
int n,m,cnt,ans=1e9+1000,data[1005];
void dfs(int now){
//	cout<<now<<endl;
	stack[++cnt]=now;visit[now]=1;
	for(int i=0;i<side[now].size();i++){
		if(cnt>=ans){cnt--;return;}
		int v=side[now][i];
		if(v==st&&cnt<ans){
				ans=cnt;
				for(int j=cnt;j>=1;j--){data[j]=stack[j];if(stack[j]==v)break;}
				cnt--;
				return;
		}
		if(visit[v])continue;
		dfs(v);
	}
	cnt--;
}
int main(){
	ios::sync_with_stdio(0);
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		int u,v;cin>>u>>v;
		side[u].push_back(v);
	}
	for(int i=1;i<=n;i++){
		cnt=0;memset(visit,0,sizeof(visit));
		st=i;dfs(i);
	}
	if(ans<1e9){
		cout<<ans<<endl;
		for(int i=1;i<=ans;i++)cout<<data[i]<<endl;
		return 0;	
	}
	cout<<"-1";
}
```