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;
#define INF 1000000001
using vi=vector<int>;
using P=pair<int,vi>;
struct node{
    vi to;
};
vi ans;
void bfs(int start, vector<node> &nodes){
    vector<bool> visited(nodes.size(),false);
    queue<P> q;
    q.push(P(start,vi{start}));
    while(q.size()){
        int v=q.front().first;
        vi path=q.front().second;
        q.pop();
        visited[v]=true;
        for (int i = 0; i < nodes[v].to.size(); i++){
            int v_to=nodes[v].to[i];
            vi tmp_path=path;
            if(v_to!=start){
                if(visited[v_to])continue;
                tmp_path.push_back(v_to);
                q.push(P(v_to,tmp_path));
            }else if(ans.size()==0 || path.size()<ans.size()){
                ans=path;
                return;
            }else return;
        }
    } 
}
int main() {
    int N,M;
    cin>>N>>M;
    vector<node> nodes(N);
    for (int i = 0; i < M; i++){
        int u,v;
        cin>>u>>v; u--; v--;
        nodes[u].to.push_back(v);
    }
    for (int i = 0; i < N; i++){
        bfs(i,nodes);
    }
    if(ans.size()==0)cout<<-1<<endl;
    else{
        cout<<ans.size()<<endl;
        for(int x : ans)cout<<x+1<<endl;
    }
}