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;
const int N = 1003;
int n, m, dis[N], par[N];
vector<int> adj[N], ans;
int main(int argc, char const *argv[])
{
	scanf("%d %d\n", &n, &m);
	for (int i = 0; i < m; ++i)
	{
		int u, v;
		scanf("%d %d\n", &u, &v);
		adj[u].push_back(v);
	}
	for (int s = 1; s <= n; ++s)
	{
		queue<int> q;
		memset(dis, -1, sizeof(dis));
		memset(par, -1, sizeof(par));
		q.push(s);
		dis[s] = 0;
		while (!q.empty())
		{
			int u = q.front();
			q.pop();
			for (int v: adj[u])
			{
				if (dis[v] == -1)
				{
					par[v] = u;
					dis[v] = 1 + dis[u];
					q.push(v);
				}
				else if (v == s)
				{
					int cur = u;
					vector<int> temp;
					while (cur != -1)
					{
						temp.push_back(cur);
						cur = par[cur];
					}
					if (temp.size() < ans.size() || ans.empty())
					{
						ans.clear();
						for (int node: temp)
							ans.push_back(node);
					}
				}
			}
		}
	}
	if (ans.empty())
		printf("%d\n", -1);
	else
	{
		printf("%d\n", (int)ans.size());
		for (int node: ans)
			printf("%d\n", node);
	}
	return 0;
}