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
sys.setrecursionlimit(10**9)

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

from collections import deque

inf = float("inf")
ans = inf
ss = []
def bfs(i):
    global ans, ss
    dist = [inf]*n
    prev = [None]*n
    que = deque()
    # 一個先から始める
    for to in G[i]:
        que.append(to)
        dist[to] = 1
        prev[to] = i
    while que:
        cur = que.popleft()
        if cur == i:
            break
        for to in G[cur]:
            if dist[to] != inf:
                continue
            dist[to] = dist[cur]+1
            prev[to] = cur
            que.append(to)

    if dist[i] < ans:
        ans = dist[i]
        ss = [i]
        for _ in range(ans-1):
            ss.append(prev[ss[-1]])
        #print(ss)

for i in range(n):
    bfs(i)

if ans == inf:
    print(-1)
else:
    print(ans)
    #print(ss)
    for i in reversed(ss):
        print(i+1)