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 cycle_detection(V, E, s):
    prev = [-1] * len(V)
    stack = [s]
    while stack:
        v = stack.pop()
        for u in E[v]:
            if u == s:
                prev[u] = v
                return (s, prev)
            if prev[u] == -1:
                stack.append(u)
                prev[u] = v
    return (-1, prev)

N, M = map(int, input().split())
V = list(range(N))
E = [[] for _ in range(N)]
edges = []
for _ in range(M):
    a, b = map(int, input().split())
    E[a-1].append(b-1)
    edges.append((a-1, b-1))

dag = False
for v in range(N):
    s, prev = cycle_detection(V, E, v)
    if s != -1: break
else:
    dag = True

if dag: print(-1)
else:
    # construct a directed cycle
    cycle = set()
    cycle.add(s)
    pv = prev[s]
    while pv != s:
        cycle.add(pv)
        pv = prev[pv]
    # shrink the directed cycle
    for a, b in edges:
        if a in cycle and b in cycle and prev[b] != a:
            pv = prev[b]
            while pv != a:
                cycle.remove(pv)
                pv = prev[pv]
            prev[b] = a
    print(len(cycle))
    print(*[v + 1 for v in cycle], sep='\n')