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 = list(map(int,input().split()))

e_list = [[] for i in range(N)]
es = []
for i in range(M):
    a,b = list(map(int,input().split()))
    a,b = a-1,b-1
    es.append((a,b))
    e_list[a].append(b)

from collections import deque
flag=False
min_d = 10**27
for vg,vi in es:
    Q = deque([vi])
    checked_list = [False for i in range(N)]
    checked_list[vi]=True
    min_path_list = [10**27 for i in range(N)] #問題によりここを変える
    min_path_list[vi] = 0
    prev_list = [-1]*N
    while len(Q)>0:
        v = Q.pop()
        for v1 in e_list[v]:
            if v!=vg or v1!=vi:
                if not checked_list[v1]:
                    checked_list[v1]=True
                    Q.appendleft(v1)
                    prev_list[v1]=v
                    min_path_list[v1]=min(min_path_list[v1],min_path_list[v]+1) #問題によりここを変える
    if min_path_list[vg]<min_d:
        flag=True
        min_vi = vi
        min_vg = vg
        min_d = min_path_list[vg]
        min_prev = prev_list
            
#print(min_d_list)
if flag:
    ans = []
    v = min_vg
    while v!=min_vi:
        ans.append(v+1)
        v=min_prev[v]
    ans.append(min_vi+1)
    ans = ans[::-1]
    
    print(len(ans))
    for a in ans:
        print(a)
else:
    print(-1)