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;
#define endl '\n'
#define pi pair<int,int>
#define ff first
#define ss second
#define boost ios_base::sync_with_stdio(false);cin.tie(nullptr)
#include "string"

const int N = 1001;
const int inf = 1e9;
int g[N][N];

void print(int n){
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++)
			cout<<g[i][j]<<" ";
		cout<<endl;
	}
}

int32_t main() { 
	boost;
	int n,m;
	cin>>n>>m;
	memset(g,1,sizeof g);
	vector<int>adj[n+1];
	for(int i=0,u,v;i<m;i++){
		cin>>u>>v;
		g[u][v] = 1;
		adj[u].push_back(v);
	}

	for(int k=1;k<=n;k++){
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++)
				g[i][j] = min(g[i][j],g[i][k]+g[k][j]);
		}
	}
	int mx = g[1][1],root=1;
	for(int i=1;i<=n;i++){
		if(g[i][i]<mx){
			mx = g[i][i];
			root = i;
		}
	}

	if(mx > 2002){
		cout<<-1<<endl;
		return 0;
	}
	cout<<mx<<endl;
	for(int i=1;i<=mx;i++){
		cout<<root<<endl;
		for(int v:adj[root]){
			if(g[v][v] == mx){
				root = v;
				break;
			}
		}
	}
}