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())
edges = [[] for _ in range(n)]
for _ in range(m):
    a, b = map(int, input().split())
    edges[a - 1].append(b - 1)

def dfs(index, visited, cycle):
    visited[index] = True
    cycle.append(index)
    for i in edges[index]:
        if visited.get(i):
            return cycle[cycle.index(i):]
        ret = dfs(i, visited, cycle)
        if ret:
            return ret
    visited[index] = False
    cycle.pop(-1)
    return None

def find_local_minimum(cycle):
    n = len(cycle)
    ind = {v: i for (i, v) in zip(range(n), cycle)}
    ret = []
    i = 0
    while True:
        ret.append(i)
        nxt = 0
        for j in [ind[v] for v in edges[cycle[i]] if v in ind]:
            if j < i:
                j += n
            nxt = max(nxt, j)
        if nxt >= n:
            ret = ret[ret.index(nxt % n):]
            break
        i += 1
        while i < nxt:
            del ind[cycle[i]]
            i += 1
    return [cycle[i % n] for i in ret]

cycle = None
visited = {}
for start in range(n):
    if start not in visited:
        cycle = dfs(start, visited, [])
        if cycle:
            break

if cycle:
    ret = find_local_minimum(cycle)
    print(len(ret))
    for i in ret:
        print(i + 1)
else:
    print(-1)