Mathematical Problem

给定奇数 $n$，求出 $n$ 个正整数满足： - 都是完全平方数。 - 长度为 $n$ 且没有前导 $0$。 - 组成这 $n$ 个数的数字（$[0,9]$ 内数字）组成的**可重集**相同。 输出任意一种方案。

The mathematicians of the 31st lyceum were given the following task: You are given an odd number $ n $ , and you need to find $ n $ different numbers that are squares of integers. But it's not that simple. Each number should have a length of $ n $ (and should not have leading zeros), and the multiset of digits of all the numbers should be the same. For example, for $ \mathtt{234} $ and $ \mathtt{432} $ , and $ \mathtt{11223} $ and $ \mathtt{32211} $ , the multisets of digits are the same, but for $ \mathtt{123} $ and $ \mathtt{112233} $ , they are not. The mathematicians couldn't solve this problem. Can you?

The first line contains an integer $ t $ ( $ 1 \leq t \leq 100 $ ) — the number of test cases. The following $ t $ lines contain one odd integer $ n $ ( $ 1 \leq n \leq 99 $ ) — the number of numbers to be found and their length. It is guaranteed that the solution exists within the given constraints. It is guaranteed that the sum of $ n^2 $ does not exceed $ 10^5 $ . The numbers can be output in any order.

For each test case, you need to output $ n $ numbers of length $ n $ — the answer to the problem. If there are several answers, print any of them.

3
1
3
5

1
169
196
961
16384
31684
36481
38416
43681

Below are the squares of the numbers that are the answers for the second test case: $ \mathtt{169} $ = $ \mathtt{13}^2 $ $ \mathtt{196} $ = $ \mathtt{14}^2 $ $ \mathtt{961} $ = $ \mathtt{31}^2 $ Below are the squares of the numbers that are the answers for the third test case: $ \mathtt{16384} $ = $ \mathtt{128}^2 $ $ \mathtt{31684} $ = $ \mathtt{178}^2 $ $ \mathtt{36481} $ = $ \mathtt{191}^2 $ $ \mathtt{38416} $ = $ \mathtt{196}^2 $ $ \mathtt{43681} $ = $ \mathtt{209}^2 $