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.readline
from collections import defaultdict

def DD(arg): return defaultdict(arg)
INF = float('inf')

N,M = list(map(int,input().split()))
graph = {i:[] for i in range(N)}
for _ in range(M):
  a,b = list(map(int,input().split()))
  graph[a-1].append(b-1)
  
Q = [i for i in range(N)]
p = [None for i in range(N)]
heiro = []
while Q:
  state = [-1 for i in range(N)]
  Flag = 0
  s = Q.pop()
  if p[s] != None: continue
  p[s] = -1
  q = [s]
  while q:
    flag = 0
    x = q.pop()
    if x < 0:
      state[~x] = 0
      continue
    q.append(~x)
    for y in graph[x]:
      state[x] = 1
      if state[y] == -1:
        p[y] = x
        q.append(y)
      elif state[y] == 0:
        continue
      else:
        p[y] = x
        now = x
        while now != y:
          heiro.append(now)
          now = p[now]
        heiro.append(y)
        flag = 1
        break
    if flag:
      Flag = 1
      break
  if Flag:
    break

def graph_size(g):
  ret = 0
  for k in g:
    ret += len(g[k])
  return ret
def squeeze(l):
  global graph
  edges = []
  subgraph = {k:[] for k in l}
  data = DD(lambda:DD(int))
  i = 0
  for k in l:
    for y in graph[k]:
      if y in l:
        subgraph[k].append(y)
        edges.append([0,k,y])
        data[k][y] = i
        i += 1
  graph = subgraph
  s = l[0]
  i = data[p[s]][s]
  edges[i][0] = 1
  now = p[s]
  heiro = [s]
  while now != s:
    heiro.append(now)
    i = data[p[now]][now]
    edges[i][0] = 1
    now = p[now]
  if len(heiro)==len(edges):
    print(len(heiro))
    for x in heiro:
      print(x+1)
    exit()
  else:
    for b,k,y in edges:
      if not b:
        p[y] = k
        now = p[y]
        heiro = [y]
        while now != y:
          heiro.append(now)
          now = p[now]
        squeeze(heiro)

if not heiro:
  print(-1)
else:
  squeeze(heiro)