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
input = sys.stdin.readline

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

L = 10**14
num = -1
for n in range(N):
    q = [n]
    checked = [False]*N
    checked[n] = True
    before = [None]*N
    c = 0
    ok = False
    while q:
        c += 1
        qq = []
        for p in q:
            for np in graph[p]:
                if not checked[np]:
                    before[np] = p
                    qq.append(np)
                    checked[np] = True
                elif np == n:
                    before[np] = p
                    ok = True
                    break
        if ok: break
        q = qq
    
    if ok and c < L:
        L = c
        num = n
        ans = []
        k = n
        for _ in range(c):
            k = before[k]
            ans.append(k+1)

if num == -1:
    print(-1)
else:
    print(len(ans))
    for a in reversed(ans):
        print(a)