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>
#define ll long long
#define ld long double
#define pi pair<ll, ll>
#define mp make_pair
#define pb push_back
#define vi vector<ll>
#define pd pair<double, double>
#define vp vector<pi>
#define f(i, n) for(int i = 0; i < n; i++)
#define fo(i, a, n) for(int i = a; i < n; i++)
#define mod (1000*1000*1000+7)
#define sz(x) int((x).size())
#define all(x) x.begin(),x.end()
#define DBG(v) cerr << #v << " = " << (v) << endl;
ll INF = (1ll<<62);

using namespace std;
vi *g;

int main(int argc, char* argv[])
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    int n, m; cin >> n >> m;
	g = new vi[n+1];
	f(i, m){
		int a, b; cin >> a >> b;
		g[a].pb(b);
	}

	vi sc;
	fo(i, 1, n+1){
		vi p(n+1, 0);
		queue<int> q;
		q.push(i);
		p[i] = -1;
		vi c;
		while(!q.empty()){
			int t = q.front();
			q.pop();
			vi &adj = g[t];
			for(auto k : adj){
				if(p[k] == 0){
					p[k] = t;
					q.push(k);
				}else if(k == i){
					c.pb(k);
					while(t != i){
						c.pb(t);
						t = p[t];
					}
					if(sz(sc) == 0 || sz(c) < sz(sc)){
						sc = c;
					}
					c.clear();
					while(!q.empty()) q.pop();
					break;
				}
			}
		}
	}
	if(sz(sc)){
		cout << sz(sc) << endl;
		f(i, sz(sc)) cout << sc[i] << endl;
	}else cout << -1 << endl;

    return 0;
}