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:
n, m = map(int, input().split())

g = [[] for _ in range(1005)]

for i in range(m):
    a, b = map(int, input().split())
    g[a].append(b)

vis = [0] * 1005


def dfs(x, p):
    if vis[x]:
        return []
    p.append(x)
    vis[x] = 1
    for v in g[x]:
        if v in p:
            return p
        t = dfs(v, p)
        if len(t):
            return t
    p.pop()
    return []


tmp = 0
ans = 10 ** 10
tmpp = []
p = []
vs = [0] * 1005


def dfs2(x):
    global ans, p, tmp, tmpp
    if len(p) >= ans:
        return
    p.append(x)
    for v in g[x]:
        if v == tmp:
            ans = len(p)
            tmpp = p[:]
            continue
        if not vs[v]:
            vs[v] = 1
            dfs2(v)
    p.pop()


a = []
for i in range(1, 1 + n):
    tmp = i
    p = []
    vs = [0] * 1005
    dfs2(i)
    if len(tmpp) == ans:
        a = tmpp[:]

if not len(a):
    print(-1)
    exit(0)

print(len(a))
for i in a:
    print(i)

#
# pre = t[0]
# next = [0] * 1005
# for i in range(1, len(t)):
#     now = t[i]
#     next[pre] = now
#     pre = now
# next[t[-1]] = t[0]
# s = set()
# for i in t:
#     s.add(i)
#
# cnt = 0
# now = t[0]
# while cnt < len(s):
#     cnt += 1
#     for i in g[now]:
#         if i == next[now]:
#             continue
#         if i in s:
#             cnt = 0
#             k = next[now]
#             while k != i:
#                 s.remove(k)
#                 k = next[k]
#             next[now] = i
#     now = next[now]
#
# print(len(s))
# for i in s:
#     print(i)