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:
def dijkstra_back(s,t,n,w,cost):
    import heapq
    #sからtへの最短経路の経路復元
    prev = [s] * n #最短経路の直前の頂点
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0
    q=[]
    heapq.heapify(q)
    heapq.heappush(q,(0,s))
    while q:
        c,v=heapq.heappop(q)
        if d[v]<c:
            continue

        for i in range(len(to[v])):
            k=to[v][i]
            if d[k]>d[v]+cost[v][k]:
                d[k]=d[v]+cost[v][k]
                heapq.heappush(q,(d[k],k))
                prev[k]=v

    path = [t]
    while prev[t] != s:
        path.append(prev[t])
        prev[t] = prev[prev[t]]
    path.append(s)
    path = path[::-1]
    return path



N,M=map(int,input().split())
fr={i:set([]) for i in range(1,N+1)}
to={i:set([]) for i in range(1,N+1)}
for i in range(0,M):
    a,b=map(int,input().split())
    fr[b].add(a)
    to[a].add(b)

delete=set([])
flag=1
while flag==1:
    flag=0
    for i in range(1,N+1):
        if (not i in delete) and (delete>=fr[i] or delete>=to[i]):
            delete.add(i)
            flag=1

node=[i for i in range(1,N+1)]
save=set(node)-delete
if save:
    hen=[]
    for i in save:
        fr[i]=fr[i]&save
        to[i]=to[i]&save
        fr[i]=list(fr[i])
        to[i]=list(to[i])

    for i in save:
        for j in fr[i]:
            hen.append((j,i))

    save=list(save)

    cost=[[float('inf') for i in range(0,N+1)] for i in range(0,N+1)]
    for i in range(0,len(hen)):
        a,b=hen[i]
        cost[a][b]=1

    ansc=10**10
    ans=[]
    for i in range(len(save)):
        for j in fr[save[i]]:
            test=dijkstra_back(save[i],j,N+1,len(hen),cost)
            if len(test)<ansc:
                ansc=len(test)
                ans=test

    print(ansc)
    for i in range(0,ansc):
        print(ans[i])

else:
    print(-1)