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 MAX 1010
using namespace std;

vector<int> adj[MAX];
int visit[MAX];
int graph[MAX][MAX];

vector<int> dfs(int v, int root)
{
	vector<int> ans;
	visit[v] = 1;
	if (v != root)
	{
		for (int i : adj[v])
		{
			if (i != root && visit[i]) return ans;
		}
		if (graph[v][root])
		{
			ans.push_back(v);
			return ans;
		}
	}
	for (int i : adj[v])
	{
		vector<int> res = dfs(i, root);
		if (!res.empty())
		{
			int flag = 1;
			for (int j : res)
			{
				if (graph[v][j] && i != j)
				{
					flag = 0;
					break;
				}
			}
			if (!flag) continue;
			res.push_back(v);
			return res;
		}
	}
	visit[v] = 0;
	return ans;
}

int main(void)
{
	int N, M;
	cin >> N >> M;
	for (int i = 0; i < M; i++)
	{
		int u, v;
		cin >> u >> v;
		u--;
		v--;
		adj[u].push_back(v);
		graph[u][v] = 1;
	}
	for (int i = 0; i < N; i++)
	{
		for (int j = 0; j < N; j++) visit[j] = 0;
		vector<int> res = dfs(i, i);
		reverse(res.begin(), res.end());
		if (!res.empty())
		{
			cout << res.size() << endl;
			for (int j : res) cout << j + 1 << endl;
			return 0;
		}
	}
	cout << -1 << endl;
	return 0;
}