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:
import sys
input = lambda: sys.stdin.readline().rstrip()
inpl = lambda: list(map(int,input().split()))
N, M = inpl()
uv = [ set() for _ in range(N)]
vu = [ set() for _ in range(N)]
for _ in range(M):
    A, B = inpl()
    uv[A-1].add(B-1)
    vu[B-1].add(A-1)

EMPTY, LOOP, TREE = 0, 1, 2
ENTER, EXIT = 0, 1
status = [EMPTY] * N
for s in range(N):
    states_pool = [(s,[],ENTER)]
    while states_pool:
        u, chain, flag = states_pool.pop()
        if flag == EXIT or status[u] == TREE:
            status[u] = TREE
            continue
        elif status[u] == LOOP:
            break
        else: # ENTER, EMPTY
            status[u] = LOOP
            states_pool.append((u,chain,EXIT))
            new_chain = chain.copy()
            new_chain.append(u)
            for c in uv[u]:
                states_pool.append((c, new_chain, ENTER))
    else:
        continue
    break
else:
    print(-1)
    exit()

n = 0
while chain[n] != u:
    n += 1
chain = chain[n:]
L = len(chain)
nextnode = [None]*N
for i in range(L-1):
    nextnode[chain[i]] = chain[i+1]
nextnode[chain[L-1]] = chain[0]
pos = [None]*N
for p in range(len(chain)):
    pos[chain[p]] = p
i = 0
while i < L:
    h = 1
    c = chain[(i+1)%L]
    for j in uv[chain[i % L]]:
        p = pos[j]
        if p is None:
            continue
        else:
            if (p-i)%L > h:
                h = (p-i)%L
                c = j
    nextnode[chain[i % L]] = c
    i += 1
    while chain[i % L] != c:
        pos[chain[i % L]] = None
        i += 1

loop = [c]
k = c
while nextnode[k] != c:
    k = nextnode[k]
    loop.append(k)
print(len(loop))
for l in loop:
    print(l+1)