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
def serp(s):
  prev = [-1]*n
  q=deque([s])
  while q:
    v = q.pop()
    for nv in e[v]:
      if nv==s: prev[nv]=v; return(s,prev)
      if prev[nv]<0: q+=nv,; prev[nv] =v
  return(-1,prev)
n,m,*t=map(int,open(0).read().split())
e,ab = [[] for i in range(n)],[]
for a,b in zip(*[iter(t)]*2):
  e[a-1]+= b-1,
  ab += [(a-1,b-1)]
for v in range(n):
  v0,prev = serp(v)
  if v0>=0: break
if v0<0: print(-1); sys.exit()
c=set()
c.add(v0)
pv = prev[v0]
while pv!=v0:
  c.add(pv)
  pv=prev[pv]
for a,b in ab:
  if a in c and b in c and prev[b]!=a:
    pv = prev[b]
    while pv !=a:
      c.remove(pv)
      pv=prev[pv]
    prev[b]=a
print(len(c))
for i in c: print(i+1)