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<vector>
using namespace std;

struct edge{
    int from, to, cost;
};

bool find_negative_loop(int v, const vector<edge>& es){
    using Cost = int;

    vector<Cost> cost(v, 0);

    for(int i = 0 ; i < v ; ++i){
        for(auto e : es){
            if(cost[e.from] + e.cost < cost[e.to]){
                if(i == v - 1){
                    return true;
                }
                cost[e.to] = cost[e.from] + e.cost;
            }
        }
    }

    return false;
}

int main(){
    int n, m;
    cin >> n >> m;

    vector<edge> es;
    for(int i = 0 ; i < m ; ++i){
        int a, b;
        cin >> a >> b;
        es.push_back({a - 1, b - 1, -1});
    }

    if(find_negative_loop(n, es)){
        cout << 0 << endl;
    }else{
        cout << -1 << endl;
    }

    return 0;
}