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:
#pragma GCC optimize("Ofast,unroll-loops")
#include<bits/stdc++.h>
#define ll long long
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB push_back
#define PF push_front
#define FILL(a,b) memset(a,b,sizeof(a))
using namespace std;
int d[1005][1005],la[1005][1005];
int n;bool vis[1005];
V<int> v[1005],ans;
queue<int> q;
void bfs(int p){
	while(!q.empty())q.pop();FILL(vis,0);
	rep(i,0,v[p].size())q.push(v[p][i]),vis[v[p][i]]=1;
	int cur,u;
	while(!q.empty()){
		cur=q.front();q.pop();
		if(cur==p)RE;
		rep(i,0,v[cur].size()){
			u=v[cur][i];
			if(!vis[u]){
				vis[u]=1;
				la[p][u]=cur;d[p][u]=d[p][cur]+1;q.push(u);
			}
		}
	}
}
int main(){
	ios::sync_with_stdio(0);
    cin.tie(0);
	FILL(d,0x3f);
	int m,x,y;
	cin>>n>>m;
	FOR(i,1,m){
		cin>>x>>y;d[x][y]=1;
		v[x].PB(y);
	}
	FOR(i,1,n){
		bfs(i);
	}
	int mini=0;
	FOR(i,1,n){
		if(d[i][i]<d[mini][mini])mini=i;
	}
	if(!mini){
		cout<<-1;RE 0;
	}
	int t=mini;
	while(t){
		ans.PB(t);t=la[mini][t];
	}
	cout<<ans.size()<<'\n';
	rep(i,0,ans.size())cout<<ans[i]<<'\n';
	RE 0;
}