#include <bits/stdc++.h>
using namespace std;
#define pie 3.141592653589793238462643383279
#define int long long
#define mod 1000000007
#define INF LLONG_MAX/5
#define all(v) v.begin(),v.end()
#define P pair<int,int>
#define S second
#define F first
#define f(i,n) for(int i=0;i<n;i++)
vector<int> v[1000],vv[1000],ans;
bool bo[1000];
int mutu(int x,bool used[1000]){
	int cnt=0,vz;
	f(i,vv[x].size()){
		if(used[vv[x][i]])cnt++;
	}
	if(cnt>1)return -1;
	cnt=0;
	f(i,v[x].size()){
		if(used[v[x][i]]){
			cnt++;
			vz=v[x][i];
		}
	}
	if(cnt==1){
		ans.push_back(x);
		return vz;
	}
	else if(cnt>=2)return -1;
	f(i,v[x].size()){
		used[v[x][i]]=true;
		int z=mutu(v[x][i],used);
		if(z!=-1){
			if(z!=x&&z!=INF){
				ans.push_back(x);
				return z;
			}
			else if(z==x){
				ans.push_back(x);
				return INF;
			}
			else return z;
		}
		used[v[x][i]]=false;
	}
	return -1;
}
signed main(){
	int a,b;
	cin>>a>>b;
	f(i,b){
		int c,d;
		cin>>c>>d;
		c--,d--;
		v[c].push_back(d);
		vv[d].push_back(c);
	}
	f(i,a){
		f(j,a)bo[j]=false;
		bo[i]=true;
		if(mutu(i,bo)!=-1){
			cout<<(int)ans.size()<<endl;
			f(i,ans.size())cout<<ans[i]+1<<endl;
			return 0;
		}
	}
	puts("-1");
}
 

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
#define pie 3.141592653589793238462643383279
#define int long long
#define mod 1000000007
#define INF LLONG_MAX/5
#define all(v) v.begin(),v.end()
#define P pair<int,int>
#define S second
#define F first
#define f(i,n) for(int i=0;i<n;i++)
vector<int> v[1000],vv[1000],ans;
bool bo[1000];
int mutu(int x,bool used[1000]){
	int cnt=0,vz;
	f(i,vv[x].size()){
		if(used[vv[x][i]])cnt++;
	}
	if(cnt>1)return -1;
	cnt=0;
	f(i,v[x].size()){
		if(used[v[x][i]]){
			cnt++;
			vz=v[x][i];
		}
	}
	if(cnt==1){
		ans.push_back(x);
		return vz;
	}
	else if(cnt>=2)return -1;
	f(i,v[x].size()){
		used[v[x][i]]=true;
		int z=mutu(v[x][i],used);
		if(z!=-1){
			if(z!=x&&z!=INF){
				ans.push_back(x);
				return z;
			}
			else if(z==x){
				ans.push_back(x);
				return INF;
			}
			else return z;
		}
		used[v[x][i]]=false;
	}
	return -1;
}
signed main(){
	int a,b;
	cin>>a>>b;
	f(i,b){
		int c,d;
		cin>>c>>d;
		c--,d--;
		v[c].push_back(d);
		vv[d].push_back(c);
	}
	f(i,a){
		f(j,a)bo[j]=false;
		bo[i]=true;
		if(mutu(i,bo)!=-1){
			cout<<(int)ans.size()<<endl;
			f(i,ans.size())cout<<ans[i]+1<<endl;
			return 0;
		}
	}
	puts("-1");
}
 

```