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

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int N,M;
vector<int>visits;
vector<int>G[1000];
int Count[1000];
int used[1000];
int rev[1000];
bool dfs(int u,int from)
{
	if(Count[u])return false;
	if(u==from)return true;
	visits.push_back(u);
	bool flag=false;
	for(int v:G[u])
	{
		Count[v]++;
		flag|=used[v]==1;
	}
	if(!flag)
	{
		used[u]=1;
		for(int v:G[u])
		{
			if(used[v]||rev[v]<rev[from])continue;
			Count[v]--;
			if(dfs(v,from))return true;
			Count[v]++;
		}
	}
	used[u]=2;
	for(int v:G[u])Count[v]--;
	visits.pop_back();
	return false;
}
int main()
{
	cin>>N>>M;
	vector<int>id(N);
	for(int i=0;i<N;i++)id[i]=i;
	for(int i=0;i<M;i++)
	{
		int u,v;cin>>u>>v;
		u--,v--;
		G[u].push_back(v);
	}
	sort(id.begin(),id.end(),[](int a,int b){return G[a].size()>G[b].size();});
	for(int i=0;i<N;i++)rev[id[i]]=i;
	for(int ii=0;ii<N;ii++)
	{
		int i=id[ii];
		for(int j=0;j<N;j++)used[j]=0;
		for(int v:G[i])Count[v]++;
		visits.push_back(i);
		for(int v:G[i])
		{
			Count[v]--;
			if(dfs(v,i))
			{
				cout<<visits.size()<<endl;
				for(int id:visits)cout<<id+1<<endl;
				return 0;
			}
			Count[v]++;
		}
		for(int v:G[i])Count[v]--;
		visits.pop_back();
	}
	cout<<-1<<endl;
}