B. Luke is a Foodie

time limit per test

1 second

memory limit per test

256 megabytes

Luke likes to eat. There are n piles of food aligned in a straight line in front of him. The i-th pile contains ai 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−ai|≤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≤i≤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

The input consists of multiple test cases. The first line contains a single integer t (1≤t≤104) --- the number of test cases. The description of test cases follows.

For each test case, the first line contains two integers, n,x (1≤n≤2⋅105, 1≤x≤109) --- 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 a1,a2,...,an (1≤ai≤109).

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105.

Output

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

Example

Input

Copy

复制代码

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

Output

Copy

复制代码

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.

解题说明：此题是一道模拟题，可以采用贪心算法，对于每一堆食物 a $i$ ，它能被吃掉的 v 的取值范围是一个区间 $a\[i$ −x,a $i$ +x]。为了最小化调整次数，尽可能长时间地保持同一个 v值。

cpp 复制代码

#include <stdio.h>

int main() 
{
    int t;
    scanf("%d", &t);
    while (t--) 
    {
        int n, x;
        scanf("%d %d", &n, &x);
        int a[n];
        for (int i = 0; i < n; i++) 
        {
            scanf("%d", &a[i]);
        }
        int changes = 0;
        long long L = a[0] - x;
        long long R = a[0] + x;

        for (int i = 1; i < n; i++) 
        {
            long long nl = a[i] - x;
            long long nr = a[i] + x;

            if (nl > R || nr < L) 
            {
                changes++;
                L = nl;
                R = nr;
            }
            else
            {
                if (nl > L)
                {
                    L = nl;
                }
                if (nr < R)
                {
                    R = nr;
                }
            }
        }
        printf("%d\n", changes);
    }

    return 0;
}