#include<bits/stdc++.h>
using namespace std;
vector<int> ans;
vector<int> g[1011];
bool vis[1011];
int dist[1011];
int n,m;
bool dfs1(int x,int s) {
	if(vis[s]){
		return true;
	}
	vis[x] = 1;
	for(int u:g[x]){
		if(!vis[u]){
			if(dfs1(u,s)){
				return true;
			}
		}
	}
	return false;
}
void dfs2(int x,vector<int>& path,int s){
	path.push_back(x);
	for(int u:g[x]){
		if(u == s) {
			if(ans.size()==0||path.size()<ans.size())
				ans = path;
			path.pop_back();
			return;
		}
		if(dist[u] > dist[x] + 1) {
			dist[u] = dist[x] + 1;
			dfs2(u,path,s);
		}
	}
	path.pop_back();
}
int main()
{
	cin>>n>>m;
	for(int i=1;i<=m;++i) {
		int x,y;
		cin>>x>>y;
		g[x].push_back(y);
	}
	for(int i=1;i<=n;++i) {
		if(!vis[i]) {
			if(dfs1(i,i)){
				memset(dist,0x7f,sizeof dist);
				dist[i] = 0;
				vector<int> path;
				dfs2(i,path,i);
			}
		}
	}
	if(ans.size() == 0) {
		cout<<-1<<endl;
	}else {
		cout<<ans.size()<<endl;
		for(int x:ans) {
			cout<<x<<endl;
		}
	}
} 
### Prompt

Generate 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;
vector<int> ans;
vector<int> g[1011];
bool vis[1011];
int dist[1011];
int n,m;
bool dfs1(int x,int s) {
	if(vis[s]){
		return true;
	}
	vis[x] = 1;
	for(int u:g[x]){
		if(!vis[u]){
			if(dfs1(u,s)){
				return true;
			}
		}
	}
	return false;
}
void dfs2(int x,vector<int>& path,int s){
	path.push_back(x);
	for(int u:g[x]){
		if(u == s) {
			if(ans.size()==0||path.size()<ans.size())
				ans = path;
			path.pop_back();
			return;
		}
		if(dist[u] > dist[x] + 1) {
			dist[u] = dist[x] + 1;
			dfs2(u,path,s);
		}
	}
	path.pop_back();
}
int main()
{
	cin>>n>>m;
	for(int i=1;i<=m;++i) {
		int x,y;
		cin>>x>>y;
		g[x].push_back(y);
	}
	for(int i=1;i<=n;++i) {
		if(!vis[i]) {
			if(dfs1(i,i)){
				memset(dist,0x7f,sizeof dist);
				dist[i] = 0;
				vector<int> path;
				dfs2(i,path,i);
			}
		}
	}
	if(ans.size() == 0) {
		cout<<-1<<endl;
	}else {
		cout<<ans.size()<<endl;
		for(int x:ans) {
			cout<<x<<endl;
		}
	}
} 
```