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())
link = [[] for _ in range(n)]
ab=[]
for i in range(m):
    a,b = list(map(int,input().split()))
    ab.append([a,b])
    link[a-1].append(b-1)

from collections import deque
def best_cycle(a,b):
    a,b=itm
    a-=1
    b-=1
    Q = deque()
    Q.append([b,[b]])
    visited=[-1]*n
    visited[b]=1

    while Q:
        now,path = Q.popleft()
        for nxt in link[now]:
            if nxt==a: # goal
                return True,path + [nxt]
            if visited[nxt]!=-1:
                continue
            visited[nxt]=1
            Q.append([nxt,path + [nxt]])
    return False,[]

mn = 10**10
ans=[]
found=False

for itm in ab:
    a,b = itm
    ret,path = best_cycle(a,b)
    if ret and len(path) < mn:
        ans=path
        mn = len(path)
        found=True

if found:
    print(len(ans))
    for a in ans:
        print(a+1)
else:
    print(-1)