CF1864B.Swap and Reverse

You are given a string $s$ of length $n$ consisting of lowercase English letters, and an integer $k$. In one step you can perform any one of the two operations below:

• Pick an index $i$ ($1 \le i \le n - 2$) and swap $s_{i}$ and $s_{i+2}$.
• Pick an index $i$ ($1 \le i \le n-k+1$) and reverse the order of letters formed by the range $[i,i+k-1]$ of the string. Formally, if the string currently is equal to $s_1\ldots s_{i-1}s_is_{i+1}\ldots s_{i+k-2}s_{i+k-1}s_{i+k}\ldots s_{n-1}s_n$, change it to $s_1\ldots s_{i-1}s_{i+k-1}s_{i+k-2}\ldots s_{i+1}s_is_{i+k}\ldots s_{n-1}s_n$.

You can make as many steps as you want (possibly, zero). Your task is to find the lexicographically smallest string you can obtain after some number of steps.

A string $a$ is lexicographically smaller than a string $b$ of the same length if and only if the following holds:

• in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$.

给你一个长度为 $n$ 的字符串 $s$，它仅由小写英文字母组成，以及一个整数 $k$。每一步中，你可以执行以下两种操作之一：

• 选择一个下标 $i$（满足 $1 \le i \le n - 2$），交换 $s_{i}$ 与 $s_{i+2}$；
• 选择一个下标 $i$（满足 $1 \le i \le n-k+1$），将字符串中区间 $[i,i+k-1]$ 内的字母顺序反转。形式化地说，若当前字符串为
  $s_1\ldots s_{i-1}s_is_{i+1}\ldots s_{i+k-2}s_{i+k-1}s_{i+k}\ldots s_{n-1}s_n$，
  则将其变为
  $s_1\ldots s_{i-1}s_{i+k-1}s_{i+k-2}\ldots s_{i+1}s_is_{i+k}\ldots s_{n-1}s_n$。

你可以执行任意多步操作（包括零步）。你的任务是求出经过若干步操作后能得到的字典序最小的字符串。

当且仅当满足以下条件时，字符串 $a$ 的字典序小于长度相同的字符串 $b$：

• 在 $a$ 与 $b$ 首次出现差异的位置上，$a$ 中的字母在字母表中早于 $b$ 中对应位置的字母。

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($1 \le k \lt n \le 10^5$).

The second line of each test case contains the string $s$ of length $n$ consisting of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

每个测试包含多个测试用例。第一行包含测试用例的数量 $t$（$1 \le t \le 10^4$）。随后是各测试用例的描述。

每个测试用例的第一行包含两个整数 $n$ 和 $k$（$1 \le k \lt n \le 10^5$）。

每个测试用例的第二行包含一个长度为 $n$ 的字符串 $s$，该字符串由小写英文字母组成。

保证所有测试用例的 $n$ 之和不超过 $10^5$。

For each test case, print the lexicographically smallest string after doing some (possibly, zero) steps.

对于每个测试用例，输出经过若干次（可能为零次）操作后字典序最小的字符串。

In the first test case, we can obtain the string "aimn" using the following operations:

1. Reverse the range $[3,4]$. The string turns into "niam".
2. Swap $s_1$ and $s_3$. The string turns into "ainm".
3. Reverse the range $[3,4]$. The string turns into "aimn".

It can be proven that we cannot obtain any string lexicographically smaller than "aimn". Therefore, "aimn" is the answer.

In the second test case, we can obtain the string "aandp" using the following operations:

1. Swap $s_3$ and $s_5$. The string turns into "paadn".
2. Swap $s_1$ and $s_3$. The string turns into "aapdn".
3. Swap $s_3$ and $s_5$. The string turns into "aandp".

It can be proven that we cannot obtain any string lexicographically smaller than "aandp". Therefore, "aandp" is the answer.

在第一个测试用例中，我们可以通过以下操作得到字符串 "aimn"：

1. 翻转区间 $[3,4]$。字符串变为 "niam"。
2. 交换 $s_1$ 与 $s_3$。字符串变为 "ainm"。
3. 翻转区间 $[3,4]$。字符串变为 "aimn"。

可以证明，我们无法得到字典序比 "aimn" 更小的任何字符串。因此，"aimn" 即为答案。

在第二个测试用例中，我们可以通过以下操作得到字符串 "aandp"：

1. 交换 $s_3$ 与 $s_5$。字符串变为 "paadn"。
2. 交换 $s_1$ 与 $s_3$。字符串变为 "aapdn"。
3. 交换 $s_3$ 与 $s_5$。字符串变为 "aandp"。

可以证明，我们无法得到字典序比 "aandp" 更小的任何字符串。因此，"aandp" 即为答案。