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
N, M = map(int, input().split())
X = [[] for i in range(N*3)]
for i in range(M):
    x, y = map(int, input().split())
    X[x-1].append(y-1)

INF = 1<<100
def BFS(n, E, i0=0):
    Q = deque([a for a in E[i0]])
    D = [INF] * n
    for a in E[i0]:
        D[a] = 1
    while Q:
        x = Q.popleft()
        for c in X[x]:
            if D[c] == INF:
                D[c] = D[x] + 1
                Q.append(c)
    return D

Y = [BFS(N, X, i)[i] for i in range(N)]
# print("Y =", Y)
mi = min(Y)
if mi == INF:
    print(-1)
else:
    for i in range(N):
        if Y[i] == mi:
            s = i
            break
    Z = [-1] * N
    for i in range(N):
        for j in X[i]:
            if Y[j] == mi:
                Z[i] = j
                break
    for i in range(N):
        s = Z[s]
    print(mi)
    for i in range(mi):
        s = Z[s]
        print(s+1)