#include <bits/stdc++.h>
using namespace std;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 100000000




int main(){
	
	int N,M;
	cin>>N>>M;
	
	vector<vector<int>> E(N,vector<int>());
	
	for(int i=0;i<M;i++){
		int a,b;
		cin>>a>>b;
		a--;b--;
		E[a].push_back(b);
	}
	
	vector<vector<int>> dis(N,vector<int>(N,Inf));
	
	for(int i=0;i<N;i++){
		dis[i][i] = 0;
		queue<int> Q;
		Q.push(i);
		while(Q.size()!=0){
			int from = Q.front();
			Q.pop();
			for(int j=0;j<E[from].size();j++){
				int to = E[from][j];
				if(dis[i][to]!=Inf)continue;
				dis[i][to] = dis[i][from]+1;
				Q.push(to);
			}
		}
	}
	
	int from = -1;
	int mini = Inf;
	set<int> S;
	
	for(int i=0;i<N;i++){
		for(int j=0;j<N;j++){
			if(i==j)continue;
			int a = dis[i][j] + dis[j][i];
			if(a<mini){
				S.clear();
				mini = a;
				from = i;
				S.insert(from);
			}
			if(a==mini){
				if(from==i){
					S.insert(j);
				}
			}
		}
	}
	
	if(mini==Inf){
		cout<<-1<<endl;
	}
	else{
		cout<<mini<<endl;
		int now = from;
		while(true){
			for(int i=0;i<E[now].size();i++){
				int to = E[now][i];
				if(S.count(to)){
					S.erase(to);
					cout<<to+1<<endl;
					now = to;
					break;
				}
			}
			if(now==from)break;
		}
	}
			
	
	return 0;
}


### Prompt

Create a solution in Cpp for 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;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 100000000




int main(){
	
	int N,M;
	cin>>N>>M;
	
	vector<vector<int>> E(N,vector<int>());
	
	for(int i=0;i<M;i++){
		int a,b;
		cin>>a>>b;
		a--;b--;
		E[a].push_back(b);
	}
	
	vector<vector<int>> dis(N,vector<int>(N,Inf));
	
	for(int i=0;i<N;i++){
		dis[i][i] = 0;
		queue<int> Q;
		Q.push(i);
		while(Q.size()!=0){
			int from = Q.front();
			Q.pop();
			for(int j=0;j<E[from].size();j++){
				int to = E[from][j];
				if(dis[i][to]!=Inf)continue;
				dis[i][to] = dis[i][from]+1;
				Q.push(to);
			}
		}
	}
	
	int from = -1;
	int mini = Inf;
	set<int> S;
	
	for(int i=0;i<N;i++){
		for(int j=0;j<N;j++){
			if(i==j)continue;
			int a = dis[i][j] + dis[j][i];
			if(a<mini){
				S.clear();
				mini = a;
				from = i;
				S.insert(from);
			}
			if(a==mini){
				if(from==i){
					S.insert(j);
				}
			}
		}
	}
	
	if(mini==Inf){
		cout<<-1<<endl;
	}
	else{
		cout<<mini<<endl;
		int now = from;
		while(true){
			for(int i=0;i<E[now].size();i++){
				int to = E[now][i];
				if(S.count(to)){
					S.erase(to);
					cout<<to+1<<endl;
					now = to;
					break;
				}
			}
			if(now==from)break;
		}
	}
			
	
	return 0;
}


```