Pattern Matchers have been designed for various sorts of patterns. Mr. HKP likes to observe patterns in numbers. After completing his extensive research on the squares of numbers, he has moved on to cubes. Now he wants to know all numbers whose cube ends in 888.

Given a number k, help Mr. HKP find the kth number (indexed from 1) whose cube ends in 888.

**Problem:** Sphere Online Judge (SPOJ) – Problem EIGHTS

**Solution:** Yeah, an other number problem! And again I just looked for patterns.

for i in xrange(0, 4444):
if str(i**3)[-3:] == "888":
print i, i**3

Which gets us:

192 7077888
442 86350888
692 331373888
942 835896888
1192 1693669888
1442 2998442888
1692 4843965888
1942 7323988888
2192 10532261888
2442 14562534888
2692 19508557888
2942 25464080888
3192 32522853888
3442 40778626888
3692 50325149888
3942 61256172888
4192 73665445888
4442 87646718888

**EDIT:** Thanks to the anonymous person who found a simple connection between the numbers but can’t properly communicate. You can find the k-th number with this simple formula:

You should see that each number ends in 2. If you look closer you see that the last two digits are 42 if i is even, and 92 if i is odd. Let’s look at the other digits. We get:

1
4
6
9
11
14
16

If you do some basic arithmetics you see that:

1
4 = 1 + 3
6 = 4 + 2
9 = 6 + 3
11 = 9 + 2
14 = 11 + 3
16 = 14 + 2

You can calculate your first digits either recursively or search for a function. This takes probably some time, so I looked this sequence up. OEIS gives us the formular: a(n) = floor((5n-2)/2) for n > 2

from math import floor
def kthnumber(k):
if k > 2:
num = "%i" % int((floor( (5 * (k - 1) + 3) / 2.0 )))
else:
if k == 1:
num = "1"
elif k == 2:
num = "4"
if k % 2 == 0:
num += "42"
else:
num += "92"
return num
t = int(raw_input())
for c in xrange(0, t):
print kthnumber(int(raw_input()))