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;
int ans[1000+10];
int xl[1000+10];
int len;
int anslen=100000;
vector<int>g[1000+10]; 
bool arr[1000+10]; 
int p=1;
void dfs(int now ,int yb)
{
	if(len+1>anslen)
		return ;
	arr[now]=true;
	xl[len++]=now;
	for(int i=0;i<g[now].size();i++)
	{
		int v=g[now][i];
		if(v==yb)
		{
			if(len<anslen)
			{
				anslen=len;
				for(int k=0;k<len;k++)
					ans[k]=xl[k];
			}
		}
		if(!arr[v])
			dfs(v,yb);
	}
	len--;
}
int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	for(int i=0;i<m;i++)
	{
		int a,b;
		scanf("%d %d",&a,&b);
		g[a].push_back(b);
	}
	for(int i=1;i<=n;i++)
	{
			memset(arr,false,sizeof(arr)); 
			dfs(i,i);
	}
	if(anslen!=100000)
	{
		printf("%d\n",anslen);
		for(int i=0;i<anslen;i++)
			printf("%d\n",ans[i]);
	}
	else
		printf("-1\n");
	return 0;
 }