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 <iostream>
#include<bits/stdc++.h>
using namespace std;
vector<int>v[1000],tmp,ans;
int n,m,a,b,c,pre[1000],dist[1000],MIN,cur;
int main(void){
    cin>>n>>m;
    MIN=n+1;
    for(int i=0;i<m;i++){
        cin>>a>>b;
        a--,b--;
        v[a].push_back(b);
    }
    for(int i=0;i<n;i++){
        queue<int> que;
        for(int i=0;i<n;i++)pre[i]=-1;
        for(int i=0;i<n;i++)dist[i]=-1;
        que.push(i);
        dist[i]=1;
        while(que.size()){
            int q=que.front();
            que.pop();
            for(int x:v[q]){
                if(dist[x]==-1){
                    dist[x]=dist[q]+1;
                    pre[x]=q;
                    que.push(x);
                }
                else if(x==i){
                    while(!que.empty())que.pop();
                    tmp={i};
                    int cur=q;
                    while(cur!=i){tmp.push_back(cur),cur=pre[cur];}
                    if(MIN>tmp.size())MIN=tmp.size(),ans=tmp;
                }
            }
        }
    }
    if(MIN==n+1)cout<<-1<<endl;
    else{
        cout<<MIN<<endl;
        for(int x:ans){
            cout<<x+1<<endl;
        }
    }
}