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 solve_f(n, g):

    for i in range(n):
        parent = [-1] * n
        queue = deque([i])
        done = False
        answer = []
        # bfs
        while len(queue) > 0:
            p = queue.popleft()
            for q in g[p]:
                if q == i:
                    r_list = []
                    r = p
                    while r != -1:
                        r_list.append(r)
                        r = parent[r]
                    # check
                    check = 1
                    for r in r_list:
                        check = max(len(set(g[r]).intersection(r_list)), check)
                    if check == 1:
                        return [len(r_list)] + [r + 1 for r in r_list]
                elif parent[q] == -1:
                    parent[q] = p
                    queue.append(q)

    return [-1]


def main():
    n, m = map(int, input().split())
    g = [[] for _ in range(n)]
    for _ in range(m):
        a, b = map(int, input().split())
        g[a - 1].append(b - 1)
    res = solve_f(n, g)
    for r in res:
        print(r)


if __name__ == "__main__":
    main()