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<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=1005;
int N,M,vis[MAXN],F[MAXN];
vector<int>P[MAXN];
queue<int>Q;
int Ans[MAXN],Len;
int rt,ans=0x3f3f3f3f,tmp[MAXN+5],dep[MAXN+5];
void Print(int x){
	if(vis[x]){
		printf("%d\n",Len);
		for(int i=1;i<=Len;i++)
			printf("%d\n",Ans[i]);
		return ;
	}
	vis[x]=1;Ans[++Len]=x;Print(tmp[x]);
}
void BFS(int u,int Pt){
	memset(vis,0,sizeof(vis));
	memset(dep,0,sizeof(dep));
	while(!Q.empty())Q.pop();
	Q.push(u);dep[u]=1,vis[u]=1;
	while(!Q.empty()){
		u=Q.front();Q.pop();
		int size=P[u].size();
		int Ok=0,flag=0;
		for(int i=0;i<size;i++){
			int v=P[u][i];
			if(v==Pt){
				if(ans>dep[u])
					rt=u,ans=dep[u],memcpy(tmp,F,sizeof(F)),tmp[Pt]=u;
				continue;
			}
			if(vis[v])
				continue;
			vis[v]=1;F[v]=u,dep[v]=dep[u]+1;
			Q.push(v);
		}
	}
	return ;
}
int main(){
	scanf("%d%d",&N,&M);
	for(int i=1,x,y;i<=M;i++){
		scanf("%d%d",&x,&y);
		P[x].push_back(y);
	}
	for(int i=1;i<=N;i++)
		BFS(i,i);
	memset(vis,0,sizeof(vis));
	if(ans==0X3F3F3F3F)
		printf("-1\n");
	else Print(rt);
}