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:
// F - Pure
#include <bits/stdc++.h>

using namespace std;

int n, m;
vector<int> ans, vi, rs;
vector<vector<int> > g;

void dfs(int u, int order)
{
	if (ans.size() > 0) return;
	if (vi[u]) {
		if (rs[u]) {
			bool found = true;
			for (int i = 0; i < n; i++)
				if (rs[i] >= rs[u]) {
					int c = 0;
					for (int j = 0; j < g[i].size(); j++)
						if (rs[g[i][j]] >= rs[u]) c++;
					if (c != 1) found = false;
				}
			if (found) {
				for (int i = 0; i < rs.size(); i++)
					if (rs[i] >= rs[u]) ans.push_back(i);
			}
		}
	} else {
		vi[u] = rs[u] = order;
		for (int i = 0; i < g[u].size(); i++) dfs(g[u][i], order + 1);
		rs[u] = 0;
	}
}

int main(int argc, char *argv[])
{
	ios::sync_with_stdio(false); cin.tie(0);
	cin >> n >> m;
	g.resize(n);
	for (int i = 0; i < m; i++) {
		int u, v;
		cin >> u >> v;
		u--; v--;
		g[u].push_back(v);
	}

	vi.assign(n, 0);
	rs.assign(n, 0);
	for (int i = 0; i < n && ans.size() == 0; i++)
		if (vi[i] == 0) dfs(i, 1);

	if (ans.size() > 0) {
		cout << ans.size() << endl;
		for (int i = 0; i < ans.size(); i++) cout << (ans[i] + 1) << endl;
	} else {
		cout << -1 << endl;
	}

	return 0;
}