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 collections import deque


def shortest_cycle_path(graph, start):
    """重み無し有向グラフgraphに対して、startから出発してstartに帰るサイクルの最短経路を求める
    サイクルが存在しないときは空のリスト [] を返す
    例) start = 0 のときの最短サイクルが 0 -> 3 -> 1 -> 0 のとき、
        [0, 3, 1, 0] を返す
    """
    n = len(graph)
    dist = [-1] * n
    prev_pos = [-1] * n
    q = deque([start])
    dist[start] = 0
    while q:
        pos = q.pop()
        for next_pos in graph[pos]:
            if dist[next_pos] == -1:
                dist[next_pos] += dist[pos] + 1
                prev_pos[next_pos] = pos
                q.append(next_pos)

    # path := サイクル経路の頂点, path_len := サイクル長
    # サイクル長 = len(path) - 1 となることに注意
    path = []
    path_len = 10**9 + 7
    for goal in range(n):
        if start == goal or dist[goal] == -1:
            continue
        for next_pos in graph[goal]:
            # goalからstartへのパスが存在するとき(つまりサイクルが存在するとき)
            if next_pos == start:
                tmp_path = []
                pos = goal
                while True:
                    tmp_path.append(pos)
                    if pos == start:
                        break
                    pos = prev_pos[pos]
                if len(tmp_path) < path_len:
                    path_len = len(tmp_path)
                    path = [start] + tmp_path[0:]
    path = path[::-1]
    return path


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

len_ans = 10**9 + 7
ans = []
for i in range(n):
    tmp = shortest_cycle_path(graph, i)
    if len(tmp) != 0 and len(tmp) < len_ans:
        len_ans = len(tmp)
        ans = tmp
if ans:
    print(len(ans) - 1)
    for i in range(len(ans) - 1):
        print(ans[i] + 1)
else:
    print(-1)