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())
from collections import deque

links = [set() for _ in range(n+1)]
for _ in range(m):
    a,b = map(int, input().split())
    links[a].add(b)

def bfs(links, start):
    parent = [-1] * (n+1)
    selfdepth = 10**6
    d = deque()
    done = {start}
    d.append((start,0))
    while(d):
        edge,depth = d.popleft()
        for i in links[edge]:
            if(i == start):
                selfdepth = depth + 1
                parent[i] = edge
                return( selfdepth, parent )
            if(not i in done):
                parent[i] = edge
                d.append((i,depth+1))
                done.add(i)

    return ( selfdepth, parent )

k = 10**6
start = -1
for i in range(1,n+1):
    selfdepth, parent_tmp = bfs(links, i)
    if(k > selfdepth):
        k = selfdepth
        start = i
        parent_k = parent_tmp

if(start == -1):
    print(-1)
    exit()

print(k)
print(start)
now = parent_k[start]
cnt = 0
while(now != start):
    print(now)
    now = parent_k[now]