Problem: Spongebob is already tired trying to reason his weird actions and calculations, so he simply asked you to find all pairs of n and m, such that there are exactly x distinct squares in the table consisting of n rows and m columns. For example, in a 3 × 5 table there are 15 squares with side one, 8 squares with side two and 3 squares with side three. The total number of distinct squares in a 3 × 5 table is 15 + 8 + 3 = 26.

Input

The first line of the input contains a single integer x (1 ≤ x ≤ 1018) — the number of squares inside the tables Spongebob is interested in.

Output

First print a single integer k — the number of tables with exactly x distinct squares inside.

Then print k pairs of integers describing the tables. Print the pairs in the order of increasing n, and in case of equality — in the order of increasing m.

Examples

Input

26


Output

6
1 26
2 9
3 5
5 3
9 2
26 1


Input

2


Output

2
1 2
2 1


Input

8


Output

4
1 8
2 3
3 2
8 1

Note

In a 1 × 2 table there are 2 1 × 1 squares. So, 2 distinct squares in total.

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In a 2 × 3 table there are 6 1 × 1 squares and 2 2 × 2 squares. That is equal to 8 squares in total.

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Solution (in PYTHON):

Solution:
def coeff(n):
    return n * (n + 1) / 2

def additional(n):
    nc2 = (n + 1) / 2  # ceil
    nf2 = n / 2        # floor
    extra = 1 if n % 2 else 0
    return nf2*nf2 + (nc2 + 1) * nc2 - extra

def free_term(numbers, n):
    if n >= len(numbers):
        numbers.append(numbers[n-1] + additional(n))
    return numbers[n]

def solve(numbers, x, n):
    c = coeff(n)
    f = free_term(numbers, n - 1) + x
    if f % c:
        return 0
    return f / c

def main():
    x = int(raw_input())
    if x == 1:
        print "1\n1 1"
        return

    numbers = []
    numbers.append(0) # for n = 0
    numbers.append(1) # for n = 1

    solutions = []
    i = 1
    last_nonzero = int(1e20)
    extra = None
    while True:
        t = solve(numbers, x, i)
        if t:
            if t < i: break
            if t == i:
                extra = t
                break
            solutions.append((i, t))
            last_nonzero = t
        i += 1
        if i >= last_nonzero: break
        if i > 2000000: break # XXX

    l = len(solutions) * 2
    if extra: l += 1
    print l
    for (x1, x2) in solutions:
        print x1, x2
    if extra:
        print extra, extra
    for (x1, x2) in solutions[::-1]:
        print x2, x1

main()