Mister B and Flight to the Moon
In order to fly to the Moon Mister B just needs to solve the following problem.There is a complete indirected graph with n vertices. You need to cover it with several simple cycles of length 3 and 4 so that each edge is in exactly 2 cycles.We are sure that Mister B will solve the problem soon and will fly to the Moon. Will you?
The only line contains single integer n (3 ≤ n ≤ 300).
If there is no answer, print -1.Otherwise, in the first line print k (1 ≤ k ≤ n2) — the number of cycles in your solution.In each of the next k lines print description of one cycle in the following format: first print integer m (3 ≤ m ≤ 4) — the length of the cycle, then print m integers v1, v2, ..., vm (1 ≤ vi ≤ n) — the vertices in the cycle in the traverse order. Each edge should be in exactly two cycles.
from collections import defaultdict
DEBUG = False

def main():
    if DEBUG:
        test()
    n = int(input())
    paths = cycles(n)
    print(len(paths))
    for p in paths:
        print('%d %s' % (len(p), ' '.join([str(v) for v in p])))

def cycles(n):
    """"""
    if n % 2 == 0:
        return even(n)
    else:
        return odd(n)

def even(n):
    """"""
    assert n % 2 == 0
    cycles = [[1, 2, 3], [2, 3, 4], [3, 4, 1], [4, 1, 2]]
    for i in range(6, n + 1, 2):
        (a, b) = (i, i - 1)
        cycles += [[a, 1, b], [a, 2, b], [a, 1, b, 2]]
        for k in range(3, i - 1, 2):
            (c, d) = (k, k + 1)
            cycles += [[a, c, b, d]] * 2
    return cycles

def odd(n):
    """"""
    assert n % 2 == 1
    cycles = [[1, 2, 3]] * 2
    for i in range(5, n + 1, 2):
        (a, b) = (i, i - 1)
        cycles += [[a, b, 1]] * 2
        for k in range(2, i - 1, 2):
            (c, d) = (k, k + 1)
            cycles += [[a, c, b, d]] * 2
    return cycles

def test():
    """"""
    print('Testing...')
    for n in range(3, 300, 21):
        check(n, cycles(n))
    print('Tests pass!')

def check(n, paths):
    """"""
    vertices = set(sum(paths, list()))
    assert vertices == set(range(1, n + 1))
    counts = defaultdict(int)
    for p in paths:
        assert len(p) == 3 or len(p) == 4
        assert len(set(p)) == len(p)
        for i in range(len(p)):
            key = tuple(sorted([p[i - 1], p[i]]))
            counts[key] += 1
    for i in range(1, n + 1):
        for j in range(i + 1, n + 1):
            assert counts[i, j] == 2
if __name__ == '__main__':
    main()
f = lambda n: [(1, 2, 3)] * 2 if n == 3 else [(1, 2, 3, 4), (1, 2, 4, 3), (1, 4, 2, 3)] if n == 4 else f(n - 2) + [(n, n - 1, 1), (n, n - 1, n - 2)] + [(n, i, n - 1, i + 1) for i in range(1, n - 2)]
ans = f(int(input()))
print(len(ans))
[print(len(i), *i) for i in ans]
import sys

def solve(n):
    if n == 3:
        return [(1, 2, 3)] * 2
    if n == 4:
        return [(1, 2, 3, 4), (1, 2, 4, 3), (1, 4, 2, 3)]
    else:
        return [*solve(n - 2), (n, n - 1, 1), (n, n - 1, n - 2), *[(n, i, n - 1, i + 1) for i in range(1, n - 2)]]
answer = solve(int(sys.stdin.readline()))
sys.stdout.write(str(len(answer)) + '\n')
for ans in answer:
    sys.stdout.write('{} {}\n'.format(len(ans), ' '.join(map(str, ans))))