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:
import os
import sys
from collections import deque

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7

N, M = list(map(int, sys.stdin.readline().split()))
AB = [list(map(int, sys.stdin.readline().split())) for _ in range(M)]
graph = [[] for _ in range(N + 1)]

for a, b in AB:
    graph[a].append(b)


def cycle_size(from_v):
    """

    :param from_v:
    :return:
    """
    parents = [None] * len(graph)
    seen = [False] * len(graph)
    que = deque([(from_v, 0)])
    while que:
        v, d = que.pop()
        for u in graph[v]:
            if seen[u]:
                continue
            seen[u] = True
            que.appendleft((u, d + 1))
            parents[u] = v
            if u == from_v:
                return d + 1, parents
    return INF, parents


min_v = None
min_size = INF
min_parents = None
for v in range(1, N + 1):
    size, parents = cycle_size(v)
    if size < min_size:
        min_v = v
        min_size = size
        min_parents = parents

if min_size < INF:
    print(min_size)
    print(min_v)
    v = min_parents[min_v]
    while v != min_v:
        print(v)
        v = min_parents[v]
else:
    print(-1)