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())
vs=set()
def dfs(v,path,pvs):
    lp=l=N+1
    for u in d[v]:
        if u in pvs:
            for i, c in enumerate(path):
                if c == u:
                    break
            nlp = path[i+1:]+[u]
            nl = len(nlp)
            if nl < l:
                lp, l = nlp, nl
        if u in vs:
            continue
        vs.add(u)
        nl, nlp = dfs(u,path+[u],pvs|{u})
        if nl < l:
            lp, l = nlp, nl
    return l,lp
d = [set() for _ in range(N+1)]
for _ in range(M):
    a,b=map(int,input().split())
    d[a].add(b)
for i in range(N):
    v=i+1
    l,lp=dfs(i+1,[v],{v})
    if l<=N:
        print("\n".join(str(x) for x in[l]+lp))
        break
else:
    print(-1)