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
sys.setrecursionlimit(100000)

N, M = list(map(int, input().split()))
AB = [list(map(int, input().split())) for _ in range(M)]

R = {}
for a, b in AB:
  if a in R:
    R[a].append(b)
  else:
    R[a] = [b]

def nasu(x, e):
  D[x] = 1
  if x not in R:
    return []
  for i in R[x]:
    if i == e:
      return [i]
    if D[i] == 0:
      t = nasu(i, e)
      if t != []:
        return [i] + t
  return []

for i in range(1, N + 1):
  D = [0] * (N + 1)
  L = nasu(i, i)
  if L != []:
    break
else:
  print(-1)
  exit()

D = [0] * (N + 1)
for i in L:
  D[i] = 1

LEN = len(L)
L = L + L

i = 0
while i < LEN:
  for j in range(i + LEN - 1, i, -1):
    if D[L[j]] != 0 and L[j] in R[L[i]]:
      for k in range(j - 1, i, -1):
        D[L[k]] = 0
      i = j
      break

A = []
for i in range(LEN):
  if D[L[i]] == 1:
    A.append(L[i])

print(len(A))
for i in A:
  print(i)