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:
#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <climits>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <string>
#include <complex>

using namespace std;
using pii = pair<int, int>;
using ll = long long;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);

	int n, m;
	cin >> n >> m;
	vector<vector<int>> adj(n);
	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		adj[a - 1].push_back(b - 1);
	}
	vector<int> answer;
	for (int i = 0; i < n; i++) {
		queue<int> q;
		vector<int> parent(n, -1);
		for (int nxt : adj[i]) {
			parent[nxt] = i;
			q.push(nxt);
		}
		while (!q.empty()) {
			int now = q.front();
			q.pop();
			if (now == i) {
				vector<int> nowanswer = { now };
				now = parent[now];
				while (now != i) {
					nowanswer.push_back(now);
					now = parent[now];
				}
				if (answer.empty() || answer.size() > nowanswer.size()) answer = nowanswer;
			}
			for (int nxt : adj[now]) {
				if (parent[nxt] == -1) {
					parent[nxt] = now;
					q.push(nxt);
				}
			}
		}
	}
	if (answer.empty()) cout << -1;
	else {
		cout << answer.size() << "\n";
		for (int ans : answer) cout << ans + 1 << "\n";
	}

	return 0;
}