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+1)]
for _ in range(m):
    a, b = map(int, input().split())
    g[a].append(b)
vis = {}
vs = []
def dfs(i):
    if i in vis:
        if vis[i]:
            del vs[:vs.index(i)]
        return vis[i]
    vis[i] = True
    vs.append(i)
    for j in g[i]:
        if dfs(j):
            return True
    vis[i] = False
    vs.pop()
    return False
for i in range(n):
    if dfs(i):
        break
else:
    print(-1)
    exit()
dist = {v: i for i, v in enumerate(vs)}
vis = set()
ws = []
nxt = 0
for i in vs:
    if dist[i] < nxt:
        continue
    k = -1
    for j in g[i]:
        if j not in dist or dist[j]<dist[i] and j not in vis:
            continue
        if k == -1:
            k = j
            continue
        dj = (dist[j]-dist[i])%len(vs)
        dk = (dist[k]-dist[i])%len(vs) 
        if dj > dk:
            k, j = j, k
    nxt = dist[k]
    vis.add(i)
    ws.append(i)
    if k in vis:
        del ws[:ws.index(k)]
        break
print(len(ws))
for i in ws:
    print(i)