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())
neighbours = [[] for i in range(N)]
for i in range(M):
    A, B = map(lambda x:int(x)-1,input().split())
    neighbours[A].append(B)

visited = [False for i in range(N)]
def bfs(node):
    parents = [None for i in range(N)]
    stack = [node]
    while stack:
        item = stack.pop(0)
        orig = item
        for neighbour in neighbours[item]:
            if visited[neighbour]:
                if parents[neighbour] == None:
                    parents[neighbour] = item
                try:
                    graph = {neighbour}
                    while item != neighbour:
                        graph.add(item)
                        item = parents[item]
                        if item == orig and item != neighbour:
                            1/0
                    print(len(graph))
                    for item in graph:
                        print(item+1)
                    return True
                except:
                    pass
            if not visited[neighbour]:
                parents[neighbour] = item
                visited[neighbour] = True
                stack.append(neighbour)
            
for i in range(N):
    if not visited[i]:
        if bfs(i):
            break
else:
   print(-1)