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<int, int> pii;

int main() {
	ios::sync_with_stdio(false); cin.tie(0);
	int n, m;
	cin >> n >> m;
	vector<vector<int>> g(n+1), h(n+1);
	for (int i = 0; i < m; ++i) {
		int a, b;
		cin >> a >> b;
		g[a].push_back(b);
		h[b].push_back(a);
	}
	vector<int> bestcycle;
	for (int i = 1; i <= n; ++i) {
		vector<int> dist(n+1, -1);
		vector<int> pre(n+1, -1);
		queue<int> q;
		dist[i] = 0;
		q.push(i);
		while (!q.empty()) {
			int j = q.front();
			q.pop();
			for (int k : h[j]) {
				if (dist[k] == -1) {
					dist[k] = dist[j] + 1;
					pre[k] = j;
					q.push(k);
				}
			}
		}
		for (int j : g[i]) {
			if (dist[j] != -1 && (bestcycle.empty() || dist[j] + 1 < bestcycle.size())) {
				bestcycle.clear();
				while (j != -1) {
					bestcycle.push_back(j);
					j = pre[j];
				}
			}
		}
	}
	if (bestcycle.empty()) {
		cout << "-1\n";
	} else {
		cout << bestcycle.size() << endl;
		for (int x : bestcycle) {
			cout << x << endl;
		}
	}
}