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 <stdio.h>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

typedef vector<int> VI;
const int MV=1005;
const int ME=2005;

int n,m;

VI g[MV];
int col[MV];
bool dfs(int u)
{
	if(col[u]) return col[u]==1;
	col[u]=1;
	for(int v:g[u])
		if(dfs(v)) return true;
	col[u]=2;
	return false;
}

VI ans,nans;
int vis[MV],pre[MV];

void bfs(int st)
{
	nans.clear();
	queue<int> qu;
	for(int i=1;i<=n;i++)
		vis[i]=0;
	qu.push(st);
	vis[st]=1;
	pre[st]=0;
	int ed=0;
	while(!qu.empty())
	{
		int u=qu.front(); qu.pop();
		for(int v:g[u])
		{
			if(v==st)
			{
				ed=u;
				break;
			}
			if(vis[v]) continue;
			pre[v]=u;
			vis[v]=1;
			qu.push(v);
		}
		if(ed) break;
	}
	//cout<<st<<" "<<ed<<endl;
	while(ed)
	{
		nans.push_back(ed);
		ed=pre[ed];
	}
}

void upd()
{
	if((int)nans.size()==0) return;
	if((int)ans.size()==0||nans.size()<ans.size())
		ans=nans;
}

int main()
{
	scanf("%d%d",&n,&m);
	int u,v;
	for(int i=0;i<m;i++)
	{
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
	}
	int f=0;
	for(int i=1;i<=n;i++)
		if(!col[i]&&dfs(i))
		{
			f=1;
			break;
		}
	if(!f)
	{
		printf("-1\n");
		return 0;
	}
	for(int i=1;i<=n;i++)
	{
		bfs(i);
		upd();
	}
	printf("%d\n",ans.size());
	for(int i:ans)
		printf("%d\n",i);
	return 0;
}