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>


int main()
{
	using vi = std::vector<int>;
	using vvi = std::vector<vi>;
	
	int N, M;
	scanf("%d%d", &N, &M);
	vvi edges(N), revEdges(N);
	for (int i{}; i < M; i++)
	{
		int A, B;
		scanf("%d%d", &A, &B);
		A--; B--;
		edges[A].push_back(B);
		revEdges[B].push_back(A);
	}

	int min{1 << 30};
	std::vector<int> ans;
	for (int begin{}; begin < N; begin++)
	{
		std::vector<int> distance(N, 1ll << 30), from(N, -1);
		distance[begin] = 0;
		using pii = std::pair<int, int>;
		std::queue<pii> que;
		que.push({0, begin});
		while (!que.empty())
		{
			auto now{que.front()};
			que.pop();
			if (distance[now.second] < now.first) continue;
			for (auto& e: edges[now.second])
				if (now.first + 1 < distance[e])
				{
					distance[e] = now.first + 1;
					from[e] = now.second;
					que.push({distance[e], e});
				}
		}
		int min_i{-1};
		for (auto& e: revEdges[begin])
			if (distance[e] + 1 < min)
			{
				min_i = e;
				min = distance[e] + 1;
			}
		if (min_i < 0) continue;
		ans.clear();
		for (int now{min_i}; now != begin; now = from[now])
			ans.push_back(now);
		ans.push_back(begin);
	}
	if (min == (1 << 30)) puts("-1");
	else
	{
		printf("%d\n", ans.size());
		for (int i{(int)ans.size() - 1}; i >= 0; i--)
			printf("%d\n", ans[i] + 1);
	}

	return 0;
}