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:
INF = float('inf')
from collections import defaultdict
from heapq import heappop, heappush

class Graph(object):

    def __init__(self):
        self.graph = defaultdict(list)

    def __len__(self):
        return len(self.graph)

    def add_edge(self, From, To, cost=1):
        self.graph[From].append((To,cost))

    def get_nodes(self):
        return self.graph.keys()

class Dijkstra(object):

    def __init__(self, graph, start):
        self.g = graph.graph
        self.dist = defaultdict(lambda:INF)
        self.dist[start] = 0
        self.prev = defaultdict(lambda: None)
        self.Q = []
        heappush(self.Q,(self.dist[start], start))

        while  self.Q:
            dist_u, u = heappop(self.Q)
            if self.dist[u] < dist_u:
                continue
            for v, cost in self.g[u]:
                alt = dist_u + cost
                if self.dist[v] > alt:
                    self.dist[v] = alt
                    self.prev[v] = u
                    heappush(self.Q,(alt, v))

    def shortest_distance(self, goal):
        return self.dist[goal]

    def shortest_path(self, goal):
        path = []
        node = goal
        while node is not None:
            path.append(node)
            node = self.prev[node]

        return path[::-1]

N, M = map(int,input().split())
g = Graph()

for i in range(M):
    a,b = map(int,input().split())
    a -= 1
    b -= 1
    g.add_edge(a,b)
    g.add_edge(a,b+N)
    g.add_edge(a+N,b)

ans = INF
j = -1

for i in range(N):
    d = Dijkstra(g,i).dist
    if d[i+N] < ans:
        ans = int(d[i+N])
        j = i

if ans != INF:
    path = Dijkstra(g,j).shortest_path(j+N)[:-1]
    print(ans)
    for node in path:
        print(node+1)
else:
    print(-1)