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:
/*Lucky_Glass*/
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=1000;

int n,m,Vtmp,Nans;
vector<int> Elnk[N+3];
int Edu[N+3],Efa[N+3],Eans[N+3],Fans[N+3];
bool Evis[N+3];

void BFS(int st){
	for(int i=1;i<=n;i++) Evis[i]=false,Efa[i]=0;
	queue<int> que;
	que.push(st);
	Evis[st]=true;
	while(!que.empty()){
		int u=que.front();que.pop();
		for(auto v : Elnk[u]){
			if(Evis[v]){
				if(v==st){
					int Ncnt=0,Pnt=u;
					while(Pnt!=st) Fans[++Ncnt]=Pnt,Pnt=Efa[Pnt];
					Fans[++Ncnt]=st;
					if(Nans>Ncnt){
						Nans=Ncnt;
						for(int i=1;i<=Ncnt;i++)
							Eans[i]=Fans[i];
					}
					return;
				}
				else continue;
			}
			Efa[v]=u;
			Evis[v]=true;
			que.push(v);
		}
	}
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		Elnk[u].push_back(v);
	}
	Nans=(1<<29);
	for(int i=1;i<=n;i++){
		BFS(i);
	}
	if(Nans==(1<<29)) printf("-1\n");
	else{
		printf("%d\n",Nans);
		for(int i=1;i<=Nans;i++)
			printf("%d\n",Eans[i]);
	}
	return 0;
}