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 int long long
#define FOR(i, s, n) for (int i = (s); i < (int)(n); i++)
#define RFOR(i, s, n) for (int i = (n) - 1; i >= (int)(s); i--)
#define REP(i, n) FOR(i, 0, n)
#define RREP(i, n) RFOR(i, 0, n)
#define ALL(a) a.begin(), a.end()
constexpr long long INF = 1e18;
template<class T>inline bool CHMAX(T&a,T b){if(a<b){a=b;return true;}return false;}
template<class T>inline bool CHMIN(T&a,T b){if(a>b){a=b;return true;}return false;}

vector<int>g[100010];

signed main(){
	int N,M;
	cin >> N >> M;
	REP(i,M){
		int a,b;
		cin >> a >> b;
		a--;b--;
		g[a].push_back(b);
	}
	vector<int>ans(N+1);
	auto min_cycle = [&](int i)->vector<int>{
		vector<int>prev(N,-1);
		queue<int>que;
		que.push(i);
		vector<int>tmp;
		while(!que.empty()){
			int now = que.front();
			que.pop();
			for(auto e:g[now]){
				if(e == i){
					while(now != -1){
						tmp.push_back(now);
						now = prev[now];
					}
					return tmp;
				}else if(prev[e] != -1)continue;
				else{
					prev[e] = now;
					que.push(e);
				}
			}
		}
		return vector<int>(N+1,0);
	};
	REP(i,N){
		auto tmp = min_cycle(i);
		if(tmp.size() < ans.size())ans = tmp;
	}
	if(ans.size() == N + 1){
		cout << -1 << endl;
		return 0;
	}
	cout << ans.size() << endl;
	for(auto &e:ans)cout << e + 1 << endl;
}