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:[] for i in range(1,N+1)}
to={i:[] for i in range(1,N+1)}
cost=[[float('inf') for i in range(0,N+1)] for i in range(0,N+1)]
for i in range(0,M):
    a,b=map(int,input().split())
    cost[a][b]=1
    fr[b].append(a)
    to[a].append(b)



ansc=10**10
ans=[]
for i in range(1,N+1):
    for j in fr[i]:
        test=dijkstra_back(i,j,N+1,M,cost)
        if len(test)==2:
            if cost[i][j]!=float('inf'):
                ansc=2
                ans=test
        elif len(test)<ansc:
            ansc=len(test)
            ans=test

if ansc!=10**10:
    print(ansc)
    for i in range(0,ansc):
        print(ans[i])
else:
    print(-1)