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:
import queue

import sys
sys.setrecursionlimit(10000)

N, M = map(int, input().split())
S = set(range(N))

to = [[] for _ in range(N)]
for _ in range(M):
    a, b = [int(x)-1 for x in input().split()]
    to[a].append(b)

visit = [0] * N
def DFS(p, n):
    if visit[p] != 0:
        return visit[p]
    else:
        visit[p] = n + 1
        for i in to[p]:
            r = DFS(i, n + 1)
            if r > 0:
                return r
        visit[p] = -1
        return -1

for i in range(N):
    r = DFS(i,0)
    if r > 0:
        for j in range(N):
            if visit[j] < r:
                S.discard(j)
        for j in range(N):
            if j in S:
                for k in to[j]:
                    if k in S:
                        if visit[k] < visit[j]:
                            for l in range(N):
                                if visit[l] < visit[k] or visit[l] > visit[j]:
                                    S.discard(l)
                        else:
                            for l in range(N):
                                if visit[l] > visit[j] and visit[l] < visit[k]:
                                    S.discard(l)
        print(len(S))
        for j in list(S):
            print(j+1)
        break
    if i == N-1:
        print(-1)