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>
#define ll long long
#define ld long double
#define mk make_pair
#define fi first
#define se second
#define vll vector<ll>
#define pii pair<ll,ll>
#define vvll vector< vector<ll> >
#define pb push_back
#define sz(v) (v).size()
#define inf 1e16
#define md 1000000007
#define all(v) (v).begin(),(v).end()
#define rep(i,a,b) for(ll i=a;i<b;++i)
#define tel(a){cout<<a<<"\n";}
#define tell(a,b){cout<<a<<" | "<<b<<"\n";}
#define telll(a,b,c){cout<<a<<" | "<<b<<" | "<<c<<"\n";}
#define teln(v,n){rep(i,0,n)cout<<v[i]+1<<"\n";cout<<"\n";}
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
#define M 1000010

int main(){
	IOS;
	ll n,m;cin>>n>>m;
	vvll w(n);
	rep(i,0,m){ll a,b;cin>>a>>b;a--,b--;w[a].pb(b);}
	ll mn=1<<30,pp,v;
	vll real_ans;
	vll ans;
	for(ll start=0;start<n;++start){
		vll p(n,-1),d(n,-1);
		queue<ll>q;
		d[start]=0;
		q.push(start);
		while(!q.empty()){
			v=q.front();q.pop();
			for(auto u:w[v]){
				if(u==start&&mn>d[v]+1){
					mn=d[v]+1,pp=v;
					ans.clear();
					while(pp!=start)ans.pb(pp),pp=p[pp];ans.pb(start);
					// teln(ans,sz(ans));
					// real_ans=ans;
				}
				else if(d[u]==-1)d[u]=d[v]+1,p[u]=v,q.push(u);
			}
		}
	}
	if(sz(ans)==0){tel(-1);}
	else {tel(sz(ans));teln(ans,sz(ans));}
	return 0;
}