Captain Flint and a Long Voyage
Contest: Codeforces Round 660 (Div. 2)
Difficulty rating: 1000
Tags: greedy, math

Captain Flint and his crew keep heading to a savage shore of Byteland for several months already, drinking rum and telling stories. In such moments uncle Bogdan often remembers his nephew Denis. Today, he has told a story about how Denis helped him to come up with an interesting problem and asked the crew to solve it.

In the beginning, uncle Bogdan wrote on a board a positive integer $$$x$$$ consisting of $$$n$$$ digits. After that, he wiped out $$$x$$$ and wrote integer $$$k$$$ instead, which was the concatenation of binary representations of digits $$$x$$$ consists of (without leading zeroes). For example, let $$$x = 729$$$, then $$$k = 111101001$$$ (since $$$7 = 111$$$, $$$2 = 10$$$, $$$9 = 1001$$$).

After some time, uncle Bogdan understood that he doesn't know what to do with $$$k$$$ and asked Denis to help. Denis decided to wipe last $$$n$$$ digits of $$$k$$$ and named the new number as $$$r$$$.

As a result, Denis proposed to find such integer $$$x$$$ of length $$$n$$$ that $$$r$$$ (as number) is maximum possible. If there are multiple valid $$$x$$$ then Denis is interested in the minimum one.

All crew members, including captain Flint himself, easily solved the task. All, except cabin boy Kostya, who was too drunk to think straight. But what about you?

Note: in this task, we compare integers ($$$x$$$ or $$$k$$$) as numbers (despite what representations they are written in), so $$$729 < 1999$$$ or $$$111 < 1000$$$.

Input format:
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Next $$$t$$$ lines contain test cases — one per test case. The one and only line of each test case contains the single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the integer $$$x$$$ you need to find.

It's guaranteed that the sum of $$$n$$$ from all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output format:
For each test case, print the minimum integer $$$x$$$ of length $$$n$$$ such that obtained by Denis number $$$r$$$ is maximum possible.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
In the second test case (with $$$n = 3$$$), if uncle Bogdan had $$$x = 998$$$ then $$$k = 100110011000$$$. Denis (by wiping last $$$n = 3$$$ digits) will obtain $$$r = 100110011$$$.

It can be proved that the $$$100110011$$$ is the maximum possible $$$r$$$ Denis can obtain and $$$998$$$ is the minimum $$$x$$$ to obtain it.

Example input:
2
1
3

Example output:
8
998

Editorial:
Statement: $$$x$$$ consists of digits $$$8-9$$$.
This is so, because if $$$x$$$ contains digits $$$0-7$$$, which in their binary notation are shorter than digits $$$8-9$$$, then the number $$$k$$$ written on the board, and therefore the number $$$r$$$ (obtained by removing the last $$$n$$$ digits of the number $$$k$$$) will be shorter than if you use only the digits $$$8$$$ and $$$9$$$, which means it will not be the maximum possible.
Statement: $$$x$$$ is $$$99 \dots 988 \dots 8$$$.
Obviously, the more $$$x$$$, the more $$$k$$$ and $$$r$$$. Therefore, to maximize $$$k$$$, $$$x$$$ must be $$$99 \dots 999$$$. However, due to the fact that $$$r$$$ is $$$k$$$ without the last $$$n$$$ digits, at the end of the number $$$x$$$ it is possible to replace a certain number of $$$9$$$ digits with $$$8$$$ so that $$$r$$$ will still be the maximum possible.
Statement: the number of digits $$$8$$$ in the number $$$x$$$ of length $$$n$$$ is equal to $$$\left\lceil \frac{n}{4} \right\rceil$$$.
$$$8_{10}=1000_2$$$ and $$$9_ {10}=1001_2$$$. We can see that the binary notations of the digits $$$8$$$ and $$$9$$$ are $$$4$$$ long and differ in the last digit.
Suppose the suffix of a number $$$x$$$ consists of $$$p$$$ digits $$$8$$$. Then the maximum $$$r$$$ is achieved if at least $$$4 \cdot p - 3$$$ digits are removed from the end of $$$k$$$. By the condition of the problem, exactly $$$n$$$ digits are removed, which means $$$4 \cdot p - 3 \le n$$$ and then $$$p = \left \lfloor \frac {n + 3} {4} \right \rfloor = \left \lceil \frac {n} {4} \right \rceil$$$.
Complexity: $$$O(n)$$$.