Luke is a Foodie
Contest: CodeTON Round 2 (Div. 1 + Div. 2, Rated, Prizes!)
Difficulty rating: 1000
Tags: brute force, greedy, implementation

Luke likes to eat. There are $$$n$$$ piles of food aligned in a straight line in front of him. The $$$i$$$-th pile contains $$$a_i$$$ units of food.

Luke will walk from the $$$1$$$-st pile towards the $$$n$$$-th pile, and he wants to eat every pile of food without walking back. When Luke reaches the $$$i$$$-th pile, he can eat that pile if and only if $$$|v - a_i| \leq x$$$, where $$$x$$$ is a fixed integer, and $$$v$$$ is Luke's food affinity.

Before Luke starts to walk, he can set $$$v$$$ to any integer. Also, for each $$$i$$$ ($$$1 \leq i \leq n$$$), Luke can change his food affinity to any integer before he eats the $$$i$$$-th pile.

Find the minimum number of changes needed to eat every pile of food.

Note that the initial choice for $$$v$$$ is not considered as a change.

Input format:
The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of test cases follows.

For each test case, the first line contains two integers, $$$n, x$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq x \leq 10^9$$$) — the number of piles, and the maximum difference between the size of a pile and Luke's food affinity, such that Luke can eat the pile.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots , a_n$$$ ($$$1 \leq a_i \leq 10^9$$$).

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

Output format:
For each test case, output an integer on a separate line, which is the minimum number of changes needed.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first test case, Luke can set $$$v$$$ to $$$5$$$ before he starts to walk. And he can walk straight to eat every piles of food without changing $$$v$$$.

In the second test case, Luke can set $$$v$$$ to $$$3$$$ before he starts to walk. And he could change $$$v$$$ to $$$10$$$ before he eats the second pile. After that, he can walk straight to eat remaining food without changing $$$v$$$.

In the fourth test case, Luke can set $$$v$$$ to $$$3$$$ before he starts to walk. And he could change $$$v$$$ to $$$8$$$ before he eats the sixth pile. After that, he can walk straight to eat remaining food without changing $$$v$$$.

In the fifth test case, Luke can set $$$v$$$ to $$$4$$$ before he starts to walk. And he could change $$$v$$$ to $$$6$$$ before he eats the fourth pile. Then he could change $$$v$$$ to $$$12$$$ before he eats the seventh pile. After that, he can walk straight to eat remaining food without changing $$$v$$$.

Example input:
7
5 3
3 8 5 6 7
5 3
3 10 9 8 7
12 8
25 3 3 17 8 6 1 16 15 25 17 23
10 2
1 2 3 4 5 6 7 8 9 10
8 2
2 4 6 8 6 4 12 14
8 2
2 7 8 9 6 13 21 28
15 5
11 4 13 23 7 10 5 21 20 11 17 5 29 16 11

Example output:
0
1
2
1
2
4
6

Editorial:
For $$$a_i$$$, if $$$|v-a_i| \leq x$$$, then $$$ a_i-x \leq v \leq a_i+x$$$.
Then consider using the greedy strategy. We will only change $$$v$$$ when we cannot find any possible $$$v$$$ to satisfy the current conditions.
Therefore, we can determine the range of $$$v$$$. Initially, set $$$l=a_1-x,r=a_1+x$$$, then $$$v \in [l,r]$$$. For $$$a_i$$$, change $$$l=\max(l,a_i-x),r=\min(r,a_i+x)$$$. Once $$$l>r$$$, which means from the last time that we changed $$$v$$$ to the current $$$a_i$$$ we cannot find any possible $$$v$$$ to satisfy the all the conditions between the two positions, we should change $$$l=a_i-x,r=a_i+x$$$, and add $$$1$$$ to answer.