#!/usr/bin/env python3

# Algorithm:
# 1. Shrink the bounding box as much as possible.
#    The wasted space `t` is now at most 6*a*b <= 6e12 (these numbers can be factored quickly)
# 2. Fill the box with a tetrahedron - that leaves 4 corners where we can take things out.
# 3. Find the largest number <= t that is a product of x,y,z with x<=a, y<=b,
#    z<=c, and cut this away from one of the corners.
# 4. Repeat this 4 times (for each of the 4 corners).
#
# v=1 and (2,2,2,25) are the only edge cases.

a, b, c, v = map(int, input().split())


# Make the bounding box tight.
while 6 * (a - 1) * b * c >= v: a -= 1
while 6 * a * (b - 1) * c >= v: b -= 1
while 6 * a * b * (c - 1) >= v: c -= 1

# t <= 6*10^12
t = 6 * a * b * c - v

# Greedily find the largest possible takeouts.
r = [(0, 0, 0)] * 4
for i in range(4):
    for w in range(t, 0, -1):
        if i == 0: p = (a, b, t//a//b)
        if i == 1: p = (1, b, t//b)
        if i == 2: p = (1, 1, min(t, c))
        if i == 3: p = (1, 1, min(t, c))
        if p[0] * p[1] * p[2] > t: p = (0,0,0)
        t -= p[0] * p[1] * p[2]
        r[i] = p

# handle edge cases
if t != 0:
    if v == 1: vertices = [(0,0,0), (1,0,0), (0,1,0), (0,0,1)]
    elif (a,b,c,v) == (2,2,2,25): vertices = [(1,0,0),(0,1,0),(1,1,0),(0,0,1),(0,2,1),(2,0,1),(2,2,1),(0,0,2),(2,0,2)]
    else: assert t==0
    print(len(vertices))
    for p in vertices: print(*p)
    exit(0)

cs = [(0, 0, 0), (0, b, c), (a, 0, c), (a, b, 0)]
ps = {(a, 0, 0), (0, b, 0), (0, 0, c), (a, b, c)}
for p, q in zip(cs, r):
    for i in range(3):
        x = [*p]
        x[i] = q[i] if x[i] == 0 else x[i] - q[i]
        ps.add((*x,))
print(len(ps))
for p in ps: print(*p)