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;

int main(){
    int N, M;
    cin >> N >> M;
    vector<int> edges[1000];
    for(int i=0; i<M; i++){
        int a, b;
        cin >> a >> b;
        edges[a-1].push_back(b-1);
    }

    vector<int> ans;
    for(int s=0; s<N; s++){
        bitset<1000> visited;
        vector<int> prv(N);
        queue<int> que;
        que.push(s);
        visited[s] = 1;
        while(que.size()){
            int i = que.front(); que.pop();
            for(int j : edges[i]){
                if(j == s){
                    vector<int> res = {i};
                    while(i != s){
                        i = prv[i];
                        res.push_back(i);
                    }
                    reverse(res.begin(), res.end());
                    if(ans.size() == 0 || ans.size() > res.size()){
                        ans = res;
                    }
                    while(que.size()) que.pop();
                    break;
                }
                if(visited[j]) continue;
                visited[j] = 1;
                prv[j] = i;
                que.push(j);
            }
        }
    }
    if(ans.size()){
        cout << ans.size() << endl;
        for(int a : ans) cout << a+1 << endl;
    }else{
        cout << -1 << endl;
    }
}