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:
N, M = map(int, input().split())
tree = dict(zip(range(1, N+1), [[] for i in range(N)]))
for i in range(M):
  a, b = map(int, input().split())
  tree[a].append(b)


is_visited = [False]*(N+1)
num_visited = 0
def find_cycle(d):
  global is_visited
  global num_visited
  cycle = []
  l = 1001
  while d:
    node = d[-1]
    if tree[node]:
      child = tree[node].pop()
      if not is_visited[child]:
        is_visited[child] = True
        num_visited += 1
      if child in cycle and node in cycle:
        idn = cycle.index(node)
        idc = cycle.index(child)
        cycle = cycle[:idn+1] + cycle[idc:]
        l = len(cycle)
      if child in d:
        idx = d.index(child)
        if len(d) - idx < l:
          cycle = d[idx:]
          l = len(cycle)
      else:
        d.append(child)
    else:
      d.pop()
  return cycle


is_visited[0] = True
num_visited += 1
i = 1
while num_visited < N+1:
  while is_visited[i]:
    i += 1
  d = [i]
  is_visited[i] = True
  num_visited += 1
  cycle = find_cycle(d)
  if cycle:
    print(len(cycle))
    print(*cycle, sep='\n')
    quit()

print(-1)