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:
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1000005
#define inf 1e9
#define eps 1e-10
using namespace std;

inline int read()
{
	int x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}

int n,m,vis[maxn],pre[maxn];
vector <int> mp[maxn],ans,tmp;
queue <int> q;

inline void bfs(int nw)
{
	for(int i=1;i<=n;i++) vis[i]=0; ans.clear();
	while(!q.empty()) q.pop(); q.push(nw),vis[nw]=1;
	while(!q.empty())
	{
		int u=q.front(); q.pop(); //cout<<nw<<" "<<u<<endl;
		for(int i=0;i<mp[u].size();i++)
		{
			int v=mp[u][i];
			if(v==nw)
			{
				int x=u; ans.push_back(nw);
				while(x!=nw) ans.push_back(x),x=pre[x];
				if(ans.size()<tmp.size()||tmp.size()==0) tmp=ans;
				return;
			}
			if(!vis[v]) q.push(v),pre[v]=u,vis[v]=1;
		}
	}
}

int main()
{
	n=read(); m=read();
	for(int i=1;i<=m;i++)
	{
		int u=read(),v=read();
		mp[u].push_back(v);
	}
	for(int i=1;i<=n;i++) bfs(i);
	if(tmp.size()==0) puts("-1");
	else
	{
		cout<<tmp.size()<<endl;
		for(int i=0;i<tmp.size();i++) printf("%d\n",tmp[i]);
	}
	
	return 0;
}