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<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
using namespace std;
const int mxn = 2010;
const int mxm = 80100;
#define ms(x,y) memset(x,y,sizeof(x))
const int inf = 0x3f3f3f3f;
int x,y,n,m,vis[mxn],b[mxn],viss[mxn],dfn[mxn],ans = inf;

vector<int>path;
int head[mxn],cnt = 0 ;
struct edge
{
	int v,nx;
}e[mxm];
void add(int u,int v)
{
	e[++cnt].v = v;
	e[cnt].nx = head[u];
	head[u] = cnt;
}
void dfs(int x,int num)
{
	vis[x] = 1;
	dfn[x] = num ;
	for(int i =head[x];~i;i = e[i].nx)
	{
		int v = e[i].v;
		if(viss[v])
		{
			if(dfn[x] - dfn[v] +1 <ans)
			{
				ans = dfn[x] - dfn[v] + 1;
				int sz = path.size();
				for(int i=  1;i<=ans;i++)
				{
					b[i] = path[sz-i];
				}
			}
		}
		if(!vis[v])
		{
			viss[v] = 1;
			path.push_back(v);
			dfs(v,num+1);
		}
	}
	path.pop_back();
	viss[x] =  0 ;
	
}
int main()
{
	memset(vis,0,sizeof(vis));
	memset(viss,0,sizeof(viss));
	memset(head,-1,sizeof(head));
	cin>>n>>m;
	for(int i = 1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);
		add(x,y);
	}
	for(int i = 1;i<=n;i++)
	{
		if(!vis[i])
		{
			path.clear();
			path.push_back(i);
			viss[i] = 1;
			dfs(i,1);
		}
	}
	if(ans == inf)printf("-1\n");
	else
	{
		printf("%d\n",ans);
		for(int i  = ans ;i>= 1;i--)printf("%d\n",b[i]);
	}
	return 0;
}