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:
def dfs(N, AB):
  status = [-1] * N
  for i in range(N):
      if status[i] == 1:
          continue
      stack = [i]
      status[i] = 0
      while stack:
          print("stack:", stack)
          v = stack[-1]
          if AB[v]:
              n = AB[v].pop()
              if status[n] == -1:
                  stack.append(n)
                  status[n] = 0
              elif status[n] == 0:
                  idx = stack.index(n)
                  print("stack:", stack, "n:", n, "idx:", idx)
                  cycle = stack[idx:]
                  
                  return cycle
          else:                  
              status[v] = 1
              stack.pop()
 
  return False

def dfs_2(N, AB):
  for c in range(N):
    stack = [(c, [])]
    while stack:
      # print("stack:", stack)
      curr, visited = stack.pop()
      if curr in visited:
        # print("cycle found:", curr)
        return visited
      else:
        # print("adding in visited:", curr)
        for i in AB[curr]:
          stack.append((i, visited+[curr]))
  return False

 
 
def find_smaller_cycle(cycle, AB):
    i = 0
    while i < len(cycle):
        v = cycle[i]
        if AB[v]:
            n = AB[v].pop()
            if n in cycle:
                r = cycle.index(n)
                if i < r:
                    cycle = cycle[:i+1] + cycle[r:]
                else:
                    cycle = cycle[r:i+1]
                i = cycle.index(v)
        else:
            i += 1
            
    return cycle
 
 
N, M = [int(i) for i in input().split()]
 
AB = [[] for _ in range(N)]
for _ in range(M):
    A, B = [int(i) - 1 for i in input().split()]
    AB[A].append(B)

# print("AB:")
# for i in AB:
#   print(i)
cycle = dfs_2(N, AB)
# print("cycle:", cycle)
if not cycle:
    print(-1)
else:
    cycle = find_smaller_cycle(cycle, AB)
            
    print(len(cycle))
    for v in cycle:
        print(v + 1)