K and Medians
Contest: 2020-2021 ICPC, NERC, Southern and Volga Russian Regional Contest (Online Mirror, ICPC Rules)
Difficulty rating: 2200
Tags: constructive algorithms, greedy, math

Let's denote the median of a sequence $$$s$$$ with odd length as the value in the middle of $$$s$$$ if we sort $$$s$$$ in non-decreasing order. For example, let $$$s = [1, 2, 5, 7, 2, 3, 12]$$$. After sorting, we get sequence $$$[1, 2, 2, \underline{3}, 5, 7, 12]$$$, and the median is equal to $$$3$$$.

You have a sequence of $$$n$$$ integers $$$[1, 2, \dots, n]$$$ and an odd integer $$$k$$$.

In one step, you choose any $$$k$$$ elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).

For example, if you have a sequence $$$[1, 2, 3, 4, 5, 6, 7]$$$ (i.e. $$$n=7$$$) and $$$k = 3$$$, then the following options for the first step are possible:

- choose $$$[1, \underline{2}, 3]$$$; $$$2$$$ is their median, so it is not erased, and the resulting sequence is $$$[2, 4, 5, 6, 7]$$$;
- choose $$$[2, \underline{4}, 6]$$$; $$$4$$$ is their median, so it is not erased, and the resulting sequence is $$$[1, 3, 4, 5, 7]$$$;
- choose $$$[1, \underline{6}, 7]$$$; $$$6$$$ is their median, so it is not erased, and the resulting sequence is $$$[2, 3, 4, 5, 6]$$$;
- and several others.

You can do zero or more steps. Can you get a sequence $$$b_1$$$, $$$b_2$$$, ..., $$$b_m$$$ after several steps?

You'll be given $$$t$$$ test cases. Solve each test case independently.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$m$$$ ($$$3 \le n \le 2 \cdot 10^5$$$; $$$3 \le k \le n$$$; $$$k$$$ is odd; $$$1 \le m < n$$$) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.

The second line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_1 < b_2 < \dots < b_m \le n$$$) — the sequence you'd like to get, given in the ascending order.

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

Output format:
For each test case, print YES if you can obtain the sequence $$$b$$$ or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Constraints:
Time limit: 2.0 s
Memory limit: 512.0 MB

Note:
In the first test case, you have sequence $$$[1, 2, 3]$$$. Since $$$k = 3$$$ you have only one way to choose $$$k$$$ elements — it's to choose all elements $$$[1, \underline{2}, 3]$$$ with median $$$2$$$. That's why after erasing all chosen elements except its median you'll get sequence $$$[2]$$$. In other words, there is no way to get sequence $$$b = [1]$$$ as the result.

In the second test case, you have sequence $$$[1, 2, 3, 4, 5, 6, 7]$$$ and one of the optimal strategies is following:

1. choose $$$k = 3$$$ elements $$$[2, \underline{3}, 4]$$$ and erase them except its median; you'll get sequence $$$[1, 3, 5, 6, 7]$$$;
2. choose $$$3$$$ elements $$$[3, \underline{5}, 6]$$$ and erase them except its median; you'll get desired sequence $$$[1, 5, 7]$$$;

In the fourth test case, you have sequence $$$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]$$$. You can choose $$$k=7$$$ elements $$$[2, 4, 6, \underline{7}, 8, 10, 13]$$$ and erase them except its median to get sequence $$$b$$$.

Example input:
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12

Example output:
NO
YES
NO
YES

Editorial:
Since after each operation we erase exactly $$$k - 1$$$ element from $$$a$$$ then if $$$(n - m) \not\equiv 0 \pmod{(k - 1)}$$$ then the answer is NO.
Otherwise, if there is such element $$$b_{pos}$$$ such that there are at least $$$\frac{k - 1}{2}$$$ erased elements lower than $$$b_i$$$ and at least $$$\frac{k - 1}{2}$$$ erased elements greater that $$$b_i$$$, then answer is YES, otherwise NO.
Let's prove this criterion in two ways. From the one side, in the last step, we should choose $$$k$$$ elements and erase them excepts its median, so the median is exactly that element $$$b_{pos}$$$.
From the other side, let's prove that if there is such $$$b_{pos}$$$ then we can always choose operations to get sequence $$$b$$$. Let's make operations in such a way that in the last step we'll erase $$$b_{pos}$$$, $$$d = \frac{k - 1}{2}$$$ elements lower $$$b_{pos}$$$ and $$$d$$$ elements greater $$$b_{pos}$$$. Since, it doesn't matter which $$$d$$$ elements to take from left and right of $$$b_{pos}$$$, we will only consider number of such elements.
Let's denote $$$x$$$ and $$$y$$$ as the initial number of elements from left and right of $$$b_{pos}$$$ to erase. We know that $$$x \ge d$$$ and $$$y \ge d$$$. and we want to make both of them equal to $$$d$$$.
Let $$$x' = x - d$$$ and $$$y' = y - d$$$. We can think that we have $$$x'$$$ free elements to erase from left and $$$y'$$$ from the right. While $$$x' + y' \ge k$$$ let's take any $$$k$$$ free elements that should be erased and erase $$$k - 1$$$ of them. Then we get situation $$$0 \le x' < k$$$, $$$0 \le y' < k$$$ and $$$x' + y' < k$$$. Since $$$(n - m)$$$ is divisible by $$$(k - 1)$$$, then it means that $$$x' + y' = k - 1$$$.
Let's look at what we can do now. We should take one of "reserved" elements to participate in erasing last $$$x' + y'$$$ free elements but it shouldn't break the situation with $$$d$$$ lower elements, $$$b_{pos}$$$ and $$$d$$$ greater elements.
If $$$x' \ge y'$$$, let's take one extra element which is lower than $$$b_{pos}$$$, then after erasing, the remaining median will also be lower than $$$b_{pos}$$$.
If $$$x' < y'$$$, let's take one extra element greater than $$$b_{pos}$$$, then the remaining median will also be greater than $$$b_{pos}$$$.
In the end, we choose $$$d$$$ elements lower, $$$b_{pos}$$$ and $$$d$$$ elements greater. Erase them except its median $$$b_{pos}$$$ and get the desired array $$$b$$$.