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>
using namespace std;
vector<int>ans;
vector<int>G[1001];
int N,M;
vector<int>vis;
bool used[1001];
void dfs(int u)
{
	vis.push_back(u);
	used[u]=true;
	for(int v:G[u])
	{
		if(vis[0]==v)
		{
			if(ans.empty()||ans.size()>vis.size())ans=vis;
		}
		else if(!used[v])dfs(v);
	}
	vis.pop_back();
}
int main()
{
	cin>>N>>M;
	for(int i=0;i<M;i++)
	{
		int u,v;cin>>u>>v;
		G[u].push_back(v);
	}
	for(int i=0;i++<N;)
	{
		for(int j=0;j++<N;)used[j]=false;
		dfs(i);
	}
	if(ans.empty())cout<<-1<<endl;
	else
	{
		cout<<ans.size()<<endl;
		for(int id:ans)cout<<id<<endl;
	}
}