#include<cstdio>
#include<stack>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1005;

int n,m,dcnt,ed;
bool vis[N];
int dep[N];
vector<int>ans;
vector<int>G[N];

bool Dfs(int u) {
	vector<int>k;
	int tmp=-1;
	for(int i=0;i<(int)G[u].size();i++) {
		int v=G[u][i];
		if(dep[v]!=-1) {
			if(tmp==-1)
				tmp=v;
			else if(dep[v]>dep[tmp])
				tmp=v;
		}
		if(!vis[v]) k.push_back(v);
		vis[v]=true;
	}
	if(tmp==-1) {
		for(int i=0;i<(int)k.size();i++) {
			int v=k[i];
			dep[v]=dep[u]+1;
			if(Dfs(v)) {
				if(dep[u]>=dep[ed])
					ans.push_back(u);
				return true;
			}
			dep[v]=-1;
		}
		for(int i=0;i<(int)k.size();i++)
			vis[k[i]]=false;
		return false;
	}
	else {
		ed=tmp;
		ans.push_back(u);
		return true;
	}
}
int main() {
	scanf("%d %d",&n,&m);
	for(int i=1,u,v;i<=m;i++) {
		scanf("%d %d",&u,&v);
		G[u].push_back(v);
	}
	memset(dep,-1,sizeof(dep));
	for(int i=1;i<=n;i++) {
		memset(vis,0,sizeof(vis));
		vis[i]=true;
		dep[i]=0;
		if(Dfs(i)) {
			printf("%d\n",(int)ans.size());
			for(int i=ans.size()-1;i>=0;i--)
				printf("%d\n",ans[i]);
			return 0;
		}
		dep[i]=-1;
	}
	printf("-1\n");
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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<cstdio>
#include<stack>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1005;

int n,m,dcnt,ed;
bool vis[N];
int dep[N];
vector<int>ans;
vector<int>G[N];

bool Dfs(int u) {
	vector<int>k;
	int tmp=-1;
	for(int i=0;i<(int)G[u].size();i++) {
		int v=G[u][i];
		if(dep[v]!=-1) {
			if(tmp==-1)
				tmp=v;
			else if(dep[v]>dep[tmp])
				tmp=v;
		}
		if(!vis[v]) k.push_back(v);
		vis[v]=true;
	}
	if(tmp==-1) {
		for(int i=0;i<(int)k.size();i++) {
			int v=k[i];
			dep[v]=dep[u]+1;
			if(Dfs(v)) {
				if(dep[u]>=dep[ed])
					ans.push_back(u);
				return true;
			}
			dep[v]=-1;
		}
		for(int i=0;i<(int)k.size();i++)
			vis[k[i]]=false;
		return false;
	}
	else {
		ed=tmp;
		ans.push_back(u);
		return true;
	}
}
int main() {
	scanf("%d %d",&n,&m);
	for(int i=1,u,v;i<=m;i++) {
		scanf("%d %d",&u,&v);
		G[u].push_back(v);
	}
	memset(dep,-1,sizeof(dep));
	for(int i=1;i<=n;i++) {
		memset(vis,0,sizeof(vis));
		vis[i]=true;
		dep[i]=0;
		if(Dfs(i)) {
			printf("%d\n",(int)ans.size());
			for(int i=ans.size()-1;i>=0;i--)
				printf("%d\n",ans[i]);
			return 0;
		}
		dep[i]=-1;
	}
	printf("-1\n");
	return 0;
}
```