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:
N,M = map(int,input().split())
AB = [list(map(int,input().split())) for _ in [0]*M]

E = [{} for _ in [0]*N]
Eb = [{} for _ in [0]*N]
for a,b in AB:
    E[a-1][b-1] = 1
    Eb[b-1][a-1] = 1

import heapq
def Dijkstra(N,E,start):
    d = [-1]*N
    q = [(0,start)]
    while q:
        ci,i = heapq.heappop(q)
        if d[i]!= -1:continue
        d[i] = ci
        for j,cj in E[i].items():
            if d[j]!=-1:continue
            heapq.heappush(q,(ci+cj,j))
    return d
INF = N*10
ans = [INF]
for i in range(N):
    d = Dijkstra(N,E,i)
    j = -1
    dj = INF
    for k in Eb[i]:
        if d[k]>0 and d[k]<dj:
            j = k
            dj = d[k]
    if (j!=-1) and ((dj+1) < ans[0]):
        ans = [dj+1,i+1]
        while j!=i:
            ans.append(j+1)
            dj = d[j]
            for k in Eb[j]:
                if d[k] == dj-1:
                    j = k
                    dj = d[k]
                    break

if ans[0] == INF : ans = [-1]
for out in ans:
    print(out)