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:
from collections import deque

class Graph(): #directed
    def __init__(self, n, edge, indexed=1):
        self.n = n
        self.graph = [[] for _ in range(n)]
        for e in edge:
            self.graph[e[0] - indexed].append(e[1] - indexed)

INF = 10**18

N, M = map(int, input().split())
edge = [tuple(map(int, input().split())) for _ in range(M)]

g = Graph(N, edge)

loopsize = INF

for i in range(N):
    root = i
    queue = deque([root])
    dist = [INF for _ in range(N)]
    dist[root] = 0
    prev = [None for _ in range(N)]
    while queue:
        node = queue.popleft()
        for adj in g.graph[node]:
            if adj == root: #閉路を検出する
                dist[root] = dist[node] + 1
                prev[root] = node
                break
            if dist[adj] != INF:
                continue
            dist[adj] = dist[node] + 1
            prev[adj] = node
            queue.append(adj)
        else:
            continue
        break
    else:
        continue
    #最小の閉路なら条件を満たす
    if loopsize > dist[root]:
        loopsize = dist[root]
        path = [root]
        node = root
        while prev[node] != root:
            node = prev[node]
            path.append(node)

if loopsize != INF:
    print(loopsize)
    for node in path:
        print(node + 1)
else:
    print(-1)