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 PYTHON3):

Solution:
import sys; input = sys.stdin.buffer.readline
from collections import defaultdict
from collections import deque
con = 10 ** 9 + 7; INF = float("inf")

def getlist():
	return list(map(int, input().split()))

class Graph(object):
	def __init__(self):
		self.graph = defaultdict(list)

	def __len__(self):
		return len(self.graph)

	def add_edge(self, a, b):
		self.graph[a].append(b)

def BFS(G, N, s, g, ans):
	Q = deque()
	Q.append(s)
	dist = [INF] * N
	dist[s] = 0
	prev = [None] * N
	jud = "No"
	while Q:
		v = Q.popleft()
		for i in G.graph[v]:
			if dist[i] == INF:
				dist[i] = dist[v] + 1
				prev[i] = v
				Q.append(i)
				if i == g:
					jud = "Yes"
					break
		if jud == "Yes":
			break

	if ans > dist[g]:
		state = g
		path = [g]
		while True:
			state = prev[state]
			path.append(state)
			if state == s:
				break
		return dist[g], path[::-1]
	else:
		return INF, []

#処理内容
def main():
	N, M = getlist()
	G = Graph()
	edgefrom = []
	edgeto = []
	for i in range(M):
		a, b = getlist()
		a -= 1; b -= 1
		edgefrom.append(a)
		edgeto.append(b)
		G.add_edge(a, b)

	ans = 100000
	path = None
	for i in range(M):
		predis, prepath = BFS(G, N, edgeto[i], edgefrom[i], ans)
		if predis != INF:
			ans = predis
			path = prepath

	if ans == 100000:
		print(-1)
	else:
		print(ans + 1)
		for i in path:
			print(i + 1)

if __name__ == '__main__':
	main()