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(10**7)
n,m=map(int,input().split())
g=[[] for _ in range(n)]
for _ in range(m):
  u,v=map(int,input().split())
  u,v=u-1,v-1
  g[u].append(v)
mi=set(range(n))
flg=False
def func(path):
  chk=set(path)
  if len(path)!=2:
    for i,v in enumerate(path):
      nvs=[nv for nv in g[v] if nv in chk]
      if len(nvs)==1:continue
      for nv in nvs:
        if nv==path[(i+1)%len(path)]:continue
        ary=[nv]
        for j,vj in enumerate(path):
          if vj==nv:break
        if i>j:
          ary=path[j:i+1]
        else:
          ary=path[:i+1]+path[j:]
        #print(path,ary,i,j)
        func(ary)
  print(len(path))
  for v in path:
    print(v+1)
  exit()

def dfs(v,seen):
  mi.discard(v)
  for nv in g[v]:
    if seen[nv]!=-1:# 巡回閉路
      seen[nv]=v
      path=[v]
      chk={v}
      while seen[v] not in chk:
        v=seen[v]
        path.append(v)
        chk.add(v)
      path.reverse()
      func(path)
    else:
      seen[nv]=v
      dfs(nv,seen)
      seen[nv]=-1
while mi:
  v=mi.pop()
  seen=[-1]*n
  dfs(v,seen)
print(-1)
exit()