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

const int maxn=1005;
int n,m,a,b,fa[maxn],flag,vis[maxn];
vector<int> to[maxn],cyc;

bool check(int x,int t)
{
	cyc.clear();
	memset(vis,0,sizeof vis);
	int now=x;
	cyc.push_back(x);
	while(now!=t&&fa[now]!=0) {
		cyc.push_back(fa[now]);
		now=fa[now];
	}
	if(cyc.size()==1) return 0;
//	for(int i=0;i<cyc.size();++i) {
//		printf("%d ",cyc[i]);
//	}
//	puts("");
	for(int i=0;i<cyc.size();++i) {
		vis[cyc[i]]=1;
	}
	for(int i=0;i<cyc.size();++i) {
		int cnt=0;
		for(int j=0;j<to[cyc[i]].size();++j) {
			if(vis[to[cyc[i]][j]]) cnt++;
		}
		if(cnt>1||cnt==0) {
			return 0;
		}
	}
	printf("%d\n",int(cyc.size()));
	for(int i=0;i<cyc.size();++i) {
		printf("%d\n",cyc[i]);
	}
	return 1;
}
	
void dfs(int x)
{
	if(flag) return ;
	for(int a=0;a<to[x].size();++a) {
		if(flag) return ;
		int t=to[x][a];
		if(fa[t]==-1) fa[t]=x,dfs(t);
		else {
			if(check(x,t)) {
				flag=1;
				return ;
			}
		}
	}
}

int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=m;++i) {
		scanf("%d %d",&a,&b);
		to[a].push_back(b);
	}
	for(int i=1;i<=n;++i) {
		memset(fa,-1,sizeof fa);
		fa[i]=0;
		dfs(i);
	}
	if(!flag) printf("-1");
}
### 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;

const int maxn=1005;
int n,m,a,b,fa[maxn],flag,vis[maxn];
vector<int> to[maxn],cyc;

bool check(int x,int t)
{
	cyc.clear();
	memset(vis,0,sizeof vis);
	int now=x;
	cyc.push_back(x);
	while(now!=t&&fa[now]!=0) {
		cyc.push_back(fa[now]);
		now=fa[now];
	}
	if(cyc.size()==1) return 0;
//	for(int i=0;i<cyc.size();++i) {
//		printf("%d ",cyc[i]);
//	}
//	puts("");
	for(int i=0;i<cyc.size();++i) {
		vis[cyc[i]]=1;
	}
	for(int i=0;i<cyc.size();++i) {
		int cnt=0;
		for(int j=0;j<to[cyc[i]].size();++j) {
			if(vis[to[cyc[i]][j]]) cnt++;
		}
		if(cnt>1||cnt==0) {
			return 0;
		}
	}
	printf("%d\n",int(cyc.size()));
	for(int i=0;i<cyc.size();++i) {
		printf("%d\n",cyc[i]);
	}
	return 1;
}
	
void dfs(int x)
{
	if(flag) return ;
	for(int a=0;a<to[x].size();++a) {
		if(flag) return ;
		int t=to[x][a];
		if(fa[t]==-1) fa[t]=x,dfs(t);
		else {
			if(check(x,t)) {
				flag=1;
				return ;
			}
		}
	}
}

int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=m;++i) {
		scanf("%d %d",&a,&b);
		to[a].push_back(b);
	}
	for(int i=1;i<=n;++i) {
		memset(fa,-1,sizeof fa);
		fa[i]=0;
		dfs(i);
	}
	if(!flag) printf("-1");
}
```