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:
import sys
from collections import deque
input = sys.stdin.readline

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

dist = [[-1] * n for _ in range(n)]
for i in range(n):
    queue = deque([(i, 0)])
    while queue:
        v, c = queue.popleft()
        for v_next in graph[v]:
            if dist[i][v_next] == -1:
                dist[i][v_next] = c + 1
                queue.append((v_next, c+1))

# print(dist)
ans_len = n + 1
for i in range(n):
    if dist[i][i] != -1:
        ans_len = min(ans_len, dist[i][i])
if ans_len == n + 1:
    print(-1)
else:
    for i in range(n):
        if dist[i][i] == ans_len:
            start_v = i
            break

    dist = [-1] * n
    prev_v = [-1] * n
    queue = deque([(start_v, 0)])
    while queue:
        v, c = queue.popleft()
        if v == start_v and c != 0:
            break
        for v_next in graph[v]:
            if dist[v_next] == -1:
                dist[v_next] = c + 1
                prev_v[v_next] = v
                queue.append((v_next, c+1))

    ans = []
    # print(start_v)
    # print(dist)
    # print(prev_v)
    v = start_v
    for i in range(ans_len):
        ans.append(v+1)
        v = prev_v[v]

    print(ans_len)
    print(*ans, sep='\n')