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>
using namespace std;
typedef long long ll;
const int MOD=1e9+7;
const int INF=1e9;
const ll LINF=1e18;

int main(){
    int n,m;
    cin>>n>>m;
    vector<int> e[n];
    for(int i=0;i<m;i++){
        int a,b;
        cin>>a>>b;
        --a,--b;
        e[a].push_back(b);
    }
    int d[n],prev[n],len=INF;
    vector<int> ans;
    for(int i=0;i<n;i++){
        fill(d,d+n,INF);
        d[i]=0,prev[i]=-1;
        queue<int> q;
        q.push(i);
        while(q.size()){
            int v=q.front();
            q.pop();
            for(int u:e[v]){
                if(u==i){
                    if(len>d[v]+1){
                        len=d[v]+1;
                        ans.clear();
                        int t=v;
                        while(t!=-1){
                            ans.push_back(t+1);
                            t=prev[t];
                        }
                        while(q.size())q.pop();
                        break;
                    }
                }
                else if(d[u]==INF){
                    d[u]=d[v]+1;
                    prev[u]=v;
                    q.push(u);
                }
            }
        }
    }
    if(ans.size()){
        cout<<ans.size()<<endl;
        for(int v:ans)cout<<v<<endl;
    }
    else cout<<-1<<endl;
}