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;

vector<int> adj[1005];
int dis[1005][1005],pa[1005][1005];
int n,m;

int get(int x)
{
	for(int i=1;i<=n;i++)
		dis[x][i]=(1<<30);
	queue<int> q;
	for(int i:adj[x])
	{
		dis[x][i]=1; pa[x][i]=x;
		q.push(i);
	}
	while(q.size())
	{
		int y=q.front(); q.pop();
		for(int i:adj[y])
		{
			if(dis[x][y]+1<dis[x][i])
			{
				dis[x][i]=dis[x][y]+1;
				pa[x][i]=y;
				q.push(i);
			}
		}
	}
	return dis[x][x];
}

int main()
{
	cin >> n >> m;
	for(int i=0;i<m;i++)
	{
		int x,y;
		cin >> x >> y;
		adj[x].push_back(y);
	}
	int ans=(1<<30),st=-1;
	for(int i=1;i<=n;i++)
	{
		int cur=get(i);
		if(cur<ans)
		{
			ans=cur;
			st=i;
		}
	}
	if(st==-1)
	{
		cout << -1 << endl;
		return 0;
	}
	get(st);
	int cur=st;
	vector<int> v;
	do{
		v.push_back(cur);
		cur=pa[st][cur];
		//cout << cur << endl;
	} while(cur != st);
	cout << v.size() << endl;
	for(int i:v)
		cout << i << endl;
}