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:
#input
N, M = map(int, input().split())
A = [0] * M
B = [0] * M
for i in range(M):
    A[i], B[i] = map(int, input().split())

#output

C = [[]] * N
for a in A:
    C[a-1] = [b-1 for i, b in enumerate(B) if A[i] == a]

#iを出発して、ループが形成できるか確認。
search_stack = [[]] * N
visited = [[]] * N
next_position = [-1] * N

for i in range(N):
    search_stack[i] = [i]
    visited[i] = []
    while search_stack[i] != []:
        next_position[i] = search_stack[i].pop(0)
        visited[i].append(next_position[i])
        for node in C[next_position[i]]:
            if node == i:
                visited[i].append(node)
                search_stack[i] = []
                break
            elif node in visited[i]:
                continue
            elif node in search_stack[i]:
                continue
            elif C[node] == []:
                continue
            else:
                search_stack[i].append(node)

short = None
for v in visited:
    if len(v) >= 2 and v[0] == v[-1]:
        if not short or len(v) < len(short):
            short = v

#結果の出力            
if short:
    del short[-1]
    print(len(short))
    for s in short:
        print(s+1)
else:
    print(-1)