from itertools import accumulate


def divisors(n: int) -> list[int]:
    """n の約数を求める"""
    s = set()
    p = 1
    while p * p <= n:
        if n % p == 0:
            s.add(p)
            s.add(n // p)
        p += 1
    return sorted(s)


def accum(a: list):
    acc = list(accumulate(a))
    return lambda l, r: acc[r] - (acc[l-1] if l > 0 else 0)


def solve_k(n: int, k: int):
    assert k > 1

    a = []
    i = 0
    while True:
        x = pow(i, k)
        if x > n: break
        a.append(x)
        i += 1

    acc = accum(a)
    ans = []
    for i in range(1, len(a)):
        if a[i] > n: break
        lo = i
        hi = len(a)-1
        while lo <= hi:
            m = (lo + hi) // 2
            x = acc(i, m)
            if x == n:
                ans.append((k, i, m))
                break
            elif x > n:
                hi = m - 1
            else:
                lo = m + 1

    return ans


def solve_k1(n: int):
    divs = divisors(2 * n)
    res = []
    for k in divs:
        a = (2*n//k - k + 1) // 2
        # 検算
        s = (a + (a+k-1)) * k // 2
        if s == n and a > 0:
            res.append((1, a, a+k-1))

    return sorted(res)


N = int(input())

ans = solve_k1(N)
for i in range(2, 41):
    if pow(2, i) > N: break
    res = solve_k(N, i)
    ans.extend(res)

print(len(ans))
for k, l, r in ans:
    print(k, l, r)