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 resolve():
    N, M = map(int, input().split())
    G = [[] for _ in range(N)]
    for _ in range(M):
        a, b = map(lambda x: int(x) - 1, input().split())
        G[a].append(b)

    shortest = N + 1
    res = []
    for s in range(N):
        dist = [-1] * N
        pre = [-1] * N
        q = deque()
        q.append(s)
        dist[s] = 0
        while q:
            v = q.popleft()
            for to in G[v]:
                if dist[to] == -1:
                    dist[to] = dist[v] + 1
                    pre[to] = v
                    q.append(to)

        for t in range(N):
            if t == s or dist[t] == -1:
                continue
            for to in G[t]:
                if to == s:  # サイクルになっている頂点
                    tmp = [s]
                    cur = t
                    while cur != s:
                        tmp.append(cur)
                        cur = pre[cur]
                    if shortest > len(tmp):
                        shortest = len(tmp)
                        res = tmp

    if shortest == N + 1:
        print(-1)
    else:
        print(len(res))
        res.sort()
        for v in res:
            print(v + 1)


if __name__ == "__main__":
    resolve()