You are given a permutation $a$ of length $n$. Recall that permutation is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order.

You have a strength of $s$ and perform $n$ moves on the permutation $a$. The $i$-th move consists of the following:

Pick two integers $x$ and $y$ such that $i \leq x \leq y \leq \min(i+s,n)$, and swap the positions of the integers $x$ and $y$ in the permutation $a$. Note that you can select $x=y$ in the operation, in which case no swap will occur.

You want to turn $a$ into another permutation $b$ after $n$ moves. However, some elements of $b$ are missing and are replaced with $-1$ instead. Count the number of ways to replace each $-1$ in $b$ with some integer from $1$ to $n$ so that $b$ is a permutation and it is possible to turn $a$ into $b$ with a strength of $s$.

Since the answer can be large, output it modulo $998\,244\,353$.


-----Input-----

The input consists of multiple test cases. The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $s$ ($1 \leq n \leq 2 \cdot 10^5$; $1 \leq s \leq n$) — the size of the permutation and your strength, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the elements of $a$. All elements of $a$ are distinct.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le n$ or $b_i = -1$) — the elements of $b$. All elements of $b$ that are not equal to $-1$ are distinct.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.


-----Output-----

For each test case, output a single integer — the number of ways to fill up the permutation $b$ so that it is possible to turn $a$ into $b$ using a strength of $s$, modulo $998\,244\,353$.


-----Examples-----

Input
6
3 1
2 1 3
3 -1 -1
3 2
2 1 3
3 -1 -1
4 1
1 4 3 2
4 3 1 2
6 4
4 2 6 3 1 5
6 1 5 -1 3 -1
7 4
1 3 6 2 7 4 5
2 5 -1 -1 -1 4 -1
14 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
Output
1
2
0
2
12
331032489


-----Note-----

In the first test case, $a=[2,1,3]$. There are two possible ways to fill out the $-1$s in $b$ to make it a permutation: $[3,1,2]$ or $[3,2,1]$. We can make $a$ into $[3,1,2]$ with a strength of $1$ as follows: $$[2,1,3] \xrightarrow[x=1,\,y=1]{} [2,1,3] \xrightarrow[x=2,\,y=3]{} [3,1,2] \xrightarrow[x=3,\,y=3]{} [3,1,2].$$ It can be proven that it is impossible to make $[2,1,3]$ into $[3,2,1]$ with a strength of $1$. Thus only one permutation $b$ satisfies the constraints, so the answer is $1$.

In the second test case, $a$ and $b$ the same as the previous test case, but we now have a strength of $2$. We can make $a$ into $[3,2,1]$ with a strength of $2$ as follows: $$[2,1,3] \xrightarrow[x=1,\,y=3]{} [2,3,1] \xrightarrow[x=2,\,y=3]{} [3,2,1] \xrightarrow[x=3,\,y=3]{} [3,2,1].$$ We can still make $a$ into $[3,1,2]$ using a strength of $1$ as shown in the previous test case, so the answer is $2$.

In the third test case, there is only one permutation $b$. It can be shown that it is impossible to turn $a$ into $b$, so the answer is $0$.

Source: codeforces | Difficulty: HARD
Tags: Sorting

Combinatorics

Amortized analysis

Complete search

Greedy algorithms

Tree algorithms


Solutions:
t = int(input())
for _ in range(t):
	(n, s) = map(int, input().split())
	a = list(map(int, input().split()))
	b = list(map(int, input().split()))
	visited = [0] * (n + 1)
	m = 1
	empty = []
	for (a_i, b_i) in zip(a, b):
		if b_i != -1 and a_i - b_i > s:
			m = 0
			break
		elif b_i == -1:
			empty.append(a_i)
		else:
			visited[b_i] += 1
	else:
		missing = [i for i in range(1, n + 1) if visited[i] == 0]
		empty.sort()
		l_max = len(missing)
		e_hi = 0
		for (e_lo, m_i) in enumerate(missing):
			while e_hi != l_max and m_i + s >= empty[e_hi]:
				e_hi += 1
			m = m * max(0, e_hi - e_lo) % 998244353
	print(m)


def convert(nums_a, nums_b, s):
	zips = list([(a, b) for (a, b) in zip(nums_a, nums_b)])
	perm = [x[1] for x in sorted(zips)]
	free_nums = set(nums_a) - set(perm)
	available_nums = set(nums_a)
	count_ar = [0] * len(perm)
	count = 1
	for (i, v) in enumerate(perm):
		rm_num = i + 1 - s - 1
		if rm_num in free_nums:
			free_nums.remove(rm_num)
		if rm_num in available_nums:
			available_nums.remove(rm_num)
		count_ar[i] = len(free_nums)
	rm_count = 0
	for (i, v) in reversed(list(enumerate(perm))):
		if perm[i] == -1:
			count *= count_ar[i] - rm_count
			count = count % 998244353
			rm_count += 1
		elif i - perm[i] >= s:
			count = 0
	print(count)
t = int(input())
for _ in range(t):
	(_, s) = list(map(int, input().split()))
	nums_a = list(map(int, input().split()))
	nums_b = list(map(int, input().split()))
	convert(nums_a, nums_b, s)