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:
from random import shuffle

N,M = map(int,input().split())

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

# find cycle
def cycle(E, visited):
    path_set = [False]*N
    path = []
    def dfs(v):
        if path_set[v]:
            path.append(v)
            return True
        if visited[v]:
            return False
        visited[v] = True
        path.append(v)
        path_set[v] = True
        for u in E[v]:
            if dfs(u):
                return True
        path.pop()
        path_set[v] = False
        return False

    while True:
        try:
            v = visited.index(False)
            if dfs(v):
                return path[path.index(path[-1]):-1]
        except ValueError:
            return None

c = cycle(E, [False]*N)
if c is None:
    print(-1)
    exit()

while True:
    c_set = set(c)
    cnt = 0
    for u in c:
        E[u] = [v for v in E[u] if v in c_set]
        shuffle(E[u])
        cnt += len(E[u])
    if cnt == len(c):
        break
    else:
        visited = [True]*N
        for v in c:
            visited[v] = False
        c = cycle(E,visited)

print(len(c))
for u in c:
    print(u+1)