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())
Edge = [list(map(int, input().split())) for i in range(m)]
adj = [[] for _ in range(n)]

for u, v in Edge:
    adj[u-1].append(v-1)

adj_bit = [0]*n
for i in range(n):
    for j in adj[i]:
        adj_bit[i] += 1 << j

for i in range(n):
    que = deque([(i, 0)])
    visited_a = [0] * n

    while que:
        v, visited = que.popleft()

        for dest in adj[v]:
            if visited_a[dest]:
                continue
            if dest == i:
                ans = [i+1]
                for j in range(1001):
                    if visited & 1:
                        ans.append(j+1)
                    visited >>= 1
                print(len(ans))
                print(*ans, sep='\n')
                exit()
            if visited & adj_bit[dest]:
                # print(i, v, dest, bin(visited), bin(adj_bit[dest]))
                continue
            visited_a[dest] = 1
            que.append((dest, visited + (1 << dest)))
else:
    print(-1)