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;
typedef pair<ll,ll> P;
typedef pair<ll,P> Q;

ll N,M;
vector<vector<ll>> G;
ll INF = 1e18;

int main(){
    cin >> N >> M;
    G.resize(N);
    for(int i=0;i<M;i++){
        ll a,b;
        cin >> a >> b;
        G[a-1].push_back(b-1);
    }
    ll ans = INF;
    vector<ll> ans_vec;
    for(int i=0;i<N;i++){
        queue<Q> q;
        vector<P> dist(N,{INF,-1});
        for(auto &e : G[i]) q.push({e,{0,-1}});
        while(!q.empty() && dist[i].first == INF){
            auto e = q.front();
            q.pop();
            ll n = e.first;
            if(dist[n].first != INF) continue;
            dist[n] = e.second;
            for(auto &f : G[n]){
                if(dist[f].first != INF) continue;
                q.push({f,{e.second.second+1,n}});
            }
        }
        if(dist[i].first == INF) continue;
        vector<ll> tmp;
        ll p = i;
        while(p != -1){
            tmp.push_back(p);
            p = dist[p].second;
        }
        if(ans > tmp.size()){
            ans_vec = tmp;
            ans = tmp.size();
        }
    }
    if(ans == INF){
        cout << -1 << endl;
    }else{
        cout << ans << endl;
        for(int i=ans-1;i>=0;i--) cout << ans_vec[i]+1 << endl;
    }
    return 0;
}