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())
AB=[list(map(int,input().split())) for i in range(M)]
c=[[] for i in range(N)]
for a,b in AB:
    c[a-1].append(b-1)
from collections import deque
import sys
sys.setrecursionlimit(10**9)
def f2(q):
    s=set(q)
    for a,b in AB:
        if a in s and b in s:
            i=q.index(b)
            if q[i-1]!=a:
                break
    if a in s and b in s and q[i-1]!=a:
        q2=deque([a,b])
        while q[(i+1)%len(q)]!=a:
            q2.append(q[(i+1)%len(q)])
            i+=1
        f2(q2)
    print(len(q),*q,sep='\n')
    exit()
v=[0]*N 
w=[0]*N
def f(p,v,q):
    v[p]=1
    w[p]=1
    q.append(p+1)
    for n in c[p]:
        if v[n]:
            f2(q)
        else:
            f(n,v,q)
    q.pop()
    v[p]=0
for p in range(N):
    if w[p]==0:
        f(p,v,deque())
print(-1)