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())
g=[[] for _ in range(n)]
for _ in range(m):
    u,v=map(int,input().split())
    u,v=u-1,v-1
    g[u].append(v)

import sys
sys.setrecursionlimit(10**7)

def func(loop):
    sl=set(loop)
    k=len(loop)
    for i in range(k):
        v=loop[i]
        for nv in g[v]:
            if nv not in sl:continue
            if nv==loop[(i+1)%k]:continue
            j=[j for j in range(k) if loop[j]==nv][0]
            if i<j:
                ary=loop[:i+1]+loop[j:]
            else:
                ary=loop[j:i+1]
            func(ary)
    print(len(loop))
    for x in loop:
        print(x+1)
    exit()

mi=set(range(n))
def dfs(v,par):
    mi.discard(v)
    for nv in g[v]:
        if par[nv]==-1:
            par[nv]=v
            dfs(nv,par)
            par[nv]=-1
        else:
            loop=[v]
            while v!=nv:
                loop.append(par[v])
                v=par[v]
            loop.reverse()
            func(loop)
while mi:
    v=mi.pop()
    dfs(v,[-1]*n)
print(-1)