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:
from collections import deque
N,M=map(int,input().split())
G=[[] for i in range(N)]
for i in range(M):
    a,b=map(int,input().split())
    G[a-1].append(b-1)
K=[0 for i in range(N)]
for i in range(N):
    for p in G[i]:
        K[p]+=1
q=deque(i for i in range(N) if K[i]==0)
res=[]
while q:
    v1=q.popleft()
    res.append(v1)
    for v2 in G[v1]:
        K[v2]-=1
        if K[v2]==0:
            q.append(v2)
if len(res)==N:
    print(-1)
    exit()
X={i for i in range(N)}
for st in range(N):
    before=[-1 for i in range(N)]
    q=deque([st])
    flag=0
    while(q):
        r=q.popleft()
        for p in G[r]:
            if before[p]==-1:
                before[p]=r
                q.append(p)
            if before[st]!=-1:
                Y={st}
                tmp=before[st]
                while(tmp!=st):
                    Y.add(tmp)
                    tmp=before[tmp]
                if len(Y)<len(X):
                    X={i for i in Y}
                flag=1
                break
        if flag:
            break
print(len(X))
for i in X:
    print(i+1)