#（別解）長さ 4 の基本パターンを連結していく方法

n = int(input())

if n == 1:
    print(-1)
    exit(0)

a2 = [1, 4]
b2 = [3, 6]
c2 = [5, 2]
a3 = [1, 2, 8]
b3 = [3, 6, 9]
c3 = [7, 5, 4]
a4 = [1, 2, 4, 11]
b4 = [7, 9, 10, 12]
c4 = [8, 5, 6, 3]
a5 = [1, 2, 3, 8, 13] # これを見つけるのがややしんどい
b5 = [7, 10, 12, 14, 15]
c5 = [11, 9, 6, 4, 5]

if n % 4 == 2:
    a = a2.copy()
    b = b2.copy()
    c = c2.copy()
elif n % 4 == 3:
    a = a3.copy()
    b = b3.copy()
    c = c3.copy()
elif n % 4 == 0:
    a = a4.copy()
    b = b4.copy()
    c = c4.copy()
else:
    a = a5.copy()
    b = b5.copy()
    c = c5.copy()

l = len(a)

for i in range(3 * l + 1, 2 * l + n + 1):
    b.append(i)

for i in range(2 * (n - l)):
    if i % 8 == 0 or i % 8 == 3 or i % 8 == 5 or i % 8 == 6:
        a.append(2 * l + n + 1 + i)
    else:
        c.append(2 * l + n + 1 + i)

# print(a)
# print(b)
# print(c)

# N = 10 の例
# [1, 4,  15, 18, 20, 21,  23, 26, 28, 29]
# [3, 6,   7,  8,  9, 10,  11, 12, 13, 14]
# [5, 2,  16, 17, 19, 22,  24, 25, 27, 30]

# l = 0
# r = 0
# for i in range(n):
#     l += a[i] * b[i]
#     r += b[i] * c[i]
# print(l, r)

print(*(a + b + c))