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:
# coding: utf-8
# Your code here!

import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline 


from collections import deque
def bfs(g,root,memo,backtrack):
    q = deque()
#    """初期化、キューに入れて訪問済みにする"""
    memo[root] = 0
    q.append(root)
    """BFS"""
    while q: #キューが空になるまで
        v = q.popleft() #キューから情報を取り出す
        for c in g[v]:
            if memo[c] > 0: continue #訪問済みならスキップ
            # 訪問済みにしてキューに入れる
            memo[c] = memo[v] + 1
            backtrack[c] = v
            q.append(c)

##############################            

n,m = [int(i) for i in readline().split()]
g = [[] for _ in range(n)]

for _ in range(m):
    a,b = [int(i)-1 for i in readline().split()]
    g[a].append(b)

res = 1000000
v0 = -1
b0 = None

for v in range(n):
    memo = [-1]*n
    backtrack = [-1]*n
    bfs(g,v,memo,backtrack)
    if 1 < memo[v] < res:
        v0 = v
        b0 = backtrack
        res = memo[v]

#print(res,v0,b0)

if res == 1000000: print(-1)
else:
    i = b0[v0]
    ans = [i+1]
    while i != v0:
        i = b0[i]
        ans.append(i+1)
    ans.reverse()
    
    print(res)
    print(*ans,sep="\n")