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<bits/stdc++.h>
using namespace std;
const int maxn=1e3+7;
const int maxm=2e3+7;
//int M[maxn][maxn];
vector<int> edge[maxn];
int main()
{
	int n,m;scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int u,v;scanf("%d%d",&u,&v);
		u--;v--;
		edge[u].push_back(v);
	}
	vector<int> ans;int mn=maxn;
	for(int s=0;s<n;s++)
	{
		vector<int> dis(maxn,-1);
		vector<int> pre(maxn,-1);
		queue<int> q;q.push(s);dis[s]=0;
		while(!q.empty())
		{
			int tmp=q.front();q.pop();
			for(int i=0;i<edge[tmp].size();i++)
			{
				int u=edge[tmp][i];
				if(dis[u]==-1)
				dis[u]=dis[tmp]+1,q.push(u),pre[u]=tmp;
			}
		}
		//for(int i=0;i<n;i++) cout<<dis[i]<<" ";
		//cout<<endl;	
		for(int t=0;t<n;t++)
		{
			if(dis[t]==-1||s==t) continue;
			vector<int> res;
			for(auto nt:edge[t])
			{
				if(nt==s)
				{
					int c=t;res.push_back(s);
					while(c!=s) res.push_back(c),c=pre[c];
					if(res.size()<mn)
					{
						mn=res.size();
						ans=res;
					}
				}
			}
		}
	}
	if(ans.size())
	{
		cout<<ans.size()<<endl;
		for(auto u:ans) cout<<u+1<<endl;
	}
	else puts("-1");
}