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
n,m = map(int,input().split())
path = [set() for _ in range(n)]
pathr= [set() for _ in range(n)]
for _ in range(m):
    a,b = map(int,input().split())
    path[a-1].add(b-1)
    pathr[b-1].add(a-1)

def bfs(start):
    que = deque()
    d = [1e100]*n
    for i in path[start]:
        que.append(i)
        d[i] = 1
    while que:
        p = que.popleft()
        for nxt in path[p]:
            if d[nxt] == 1e100:
                que.append(nxt)
            d[nxt] = min(d[nxt], d[p] + 1)
    if d[start] == 1e100:
        return [0] * (n+1)
    ret = [0]*d[start]
    ret[-1] = start
    now = start
    for i in range(d[start] - 1)[::-1]:
        for p in pathr[now]:
            if d[p] == i+1:
                now = p
                ret[i] = now
                break
    return ret
ans = [0]*(n+1)
for i in range(n):
    tmp = bfs(i)
    if len(ans) > len(tmp):
        ans = tmp
if len(ans) > n:
  print(-1)
else:
    print(len(ans))
    for i in ans:
        print(i+1)