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
from collections import deque

input = sys.stdin.readline

N, M = list(map(int,input().split()))
neighbor = [[] for i in range(N)]
for i in range(M):
	a, b = list(map(int,input().split()))
	neighbor[a-1].append(b-1)

rflag = False
for s in range(N):
	pre = [-1]*N
	start = s
	queue = [start]
	while len(queue)>0:
		nq = []
		for i in queue:
			for j in neighbor[i]:
				if pre[j]==-1:
					pre[j] = i
					nq.append(j)
					if j==start:
						rflag = True
						break
				# else:
				# 	pre[j] = i
				# 	print(pre)
				# 	break

			if rflag: break
		if rflag: break
		queue = nq
	if rflag: break

if not rflag:
	print(-1)
else:
	v = [0]*N
	v[j] = 1
	roop = [j]
	i = j
	while pre[i]!=j:
		pret = pre[i]
		roop.append(pret)
		v[pret] = 1
		i = pret
	roop = roop+[j]
	roop = roop[::-1]

	update = True
	while update:
		update = False
		for i,vi in enumerate(roop):
			if i==len(roop)-1: break
			for ni in neighbor[vi]:
				if ni!=roop[i+1] and v[ni]==1:
					pre[ni] = vi

					v = [0]*N
					v[ni] = 1
					roopn = [ni]
					j = ni
					while pre[j]!=ni:
						pret = pre[j]
						roopn.append(pret)
						v[pret] = 1
						j = pret
					roopn = roopn+[ni]
					update = True
					roop = roopn[::-1]
					break

			if update: break

	K = len(roop)-1
	print(K)
	print(*map(lambda x:x+1,roop[:-1]), sep='\n')