CF1384A.Common Prefixes

The length of the longest common prefix of two strings $s = s_1 s_2 \ldots s_n$ and $t = t_1 t_2 \ldots t_m$ is defined as the maximum integer $k$ ($0 \le k \le min(n,m)$) such that $s_1 s_2 \ldots s_k$ equals $t_1 t_2 \ldots t_k$.

Koa the Koala initially has $n+1$ strings $s_1, s_2, \dots, s_{n+1}$.

For each $i$ ($1 \le i \le n$) she calculated $a_i$ — the length of the longest common prefix of $s_i$ and $s_{i+1}$.

Several days later Koa found these numbers, but she couldn't remember the strings.

So Koa would like to find some strings $s_1, s_2, \dots, s_{n+1}$ which would have generated numbers $a_1, a_2, \dots, a_n$. Can you help her?

If there are many answers print any. We can show that answer always exists for the given constraints.

两个字符串 $s = s_1 s_2 \ldots s_n$ 和 $t = t_1 t_2 \ldots t_m$ 的最长公共前缀长度定义为满足 $s_1 s_2 \ldots s_k = t_1 t_2 \ldots t_k$ 的最大整数 $k$（其中 $0 \le k \le \min(n,m)$）。

考拉熊 Koa 最初有 $n+1$ 个字符串 $s_1, s_2, \dots, s_{n+1}$。

对每个 $i$（$1 \le i \le n$），她计算了 $a_i$ —— 即 $s_i$ 与 $s_{i+1}$ 的最长公共前缀长度。

几天后，Koa 找到了这些数值 $a_i$，但她已记不得原来的字符串了。

因此，Koa 希望找出一组字符串 $s_1, s_2, \dots, s_{n+1}$，使得它们能生成给定的数值 $a_1, a_2, \dots, a_n$。你能帮她吗？

若存在多个解，输出任意一个即可。可以证明：在题目给定的约束条件下，解一定存在。

Each test contains multiple test cases. The first line contains $t$ ($1 \le t \le 100$) — the number of test cases. Description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the number of elements in the list $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 50$) — the elements of $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

每个测试包含多个测试用例。第一行包含 $t$（$1 \le t \le 100$）—— 测试用例的数量。随后是各测试用例的描述。

每个测试用例的第一行包含一个整数 $n$（$1 \le n \le 100$）—— 列表 $a$ 中的元素个数。

每个测试用例的第二行包含 $n$ 个整数 $a_1, a_2, \ldots, a_n$（$0 \le a_i \le 50$）—— $a$ 的元素。

保证所有测试用例的 $n$ 之和不超过 $100$。

For each test case:

Output $n+1$ lines. In the $i$-th line print string $s_i$ ($1 \le |s_i| \le 200$), consisting of lowercase Latin letters. Length of the longest common prefix of strings $s_i$ and $s_{i+1}$ has to be equal to $a_i$.

If there are many answers print any. We can show that answer always exists for the given constraints.

对于每个测试用例：

输出 $n+1$ 行。在第 $i$ 行输出字符串 $s_i$（$1 \le |s_i| \le 200$），该字符串仅由小写拉丁字母组成。字符串 $s_i$ 与 $s_{i+1}$ 的最长公共前缀长度必须等于 $a_i$。

若存在多个合法答案，输出任意一个即可。可以证明：在给定约束条件下，答案一定存在。

In the $1$-st test case one of the possible answers is $s = [aeren, ari, arousal, around, ari]$.

Lengths of longest common prefixes are:

• Between $\color{red}{a}eren$ and $\color{red}{a}ri$ $\rightarrow 1$
• Between $\color{red}{ar}i$ and $\color{red}{ar}ousal$ $\rightarrow 2$
• Between $\color{red}{arou}sal$ and $\color{red}{arou}nd$ $\rightarrow 4$
• Between $\color{red}{ar}ound$ and $\color{red}{ar}i$ $\rightarrow 2$

在第 $1$ 个测试用例中，一种可能的答案是 $s = [aeren, ari, arousal, around, ari]$。

最长公共前缀的长度分别为：

• $\color{red}{a}eren$ 与 $\color{red}{a}ri$ 之间 $\rightarrow 1$
• $\color{red}{ar}i$ 与 $\color{red}{ar}ousal$ 之间 $\rightarrow 2$
• $\color{red}{arou}sal$ 与 $\color{red}{arou}nd$ 之间 $\rightarrow 4$
• $\color{red}{ar}ound$ 与 $\color{red}{ar}i$ 之间 $\rightarrow 2$