CF1919C.Grouping Increases

You are given an array $a$ of size $n$ . You will do the following process to calculate your penalty:

1. Split array $a$ into two (possibly empty) subsequences $^\dagger$ $s$ and $t$ such that every element of $a$ is either in $s$ or $t^\ddagger$ .
2. For an array $b$ of size $m$ , define the penalty $p(b)$ of an array $b$ as the number of indices $i$ between $1$ and $m - 1$ where $b_i < b_{i + 1}$ .
3. The total penalty you will receive is $p(s) + p(t)$ .

If you perform the above process optimally, find the minimum possible penalty you will receive.

$^\dagger$ A sequence $x$ is a subsequence of a sequence $y$ if $x$ can be obtained from $y$ by the deletion of several (possibly, zero or all) elements.

$^\ddagger$ Some valid ways to split array $a=[3,1,4,1,5]$ into $(s,t)$ are $([3,4,1,5],[1])$ , $([1,1],[3,4,5])$ and $([\,],[3,1,4,1,5])$ while some invalid ways to split $a$ are $([3,4,5],[1])$ , $([3,1,4,1],[1,5])$ and $([1,3,4],[5,1])$ .

Each test contains 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 the test cases follows.

The first line of each test case contains a single integer $n$ ( $1\le n\le 2\cdot 10^5$ ) — the size of the array $a$ .

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ( $1 \le a_i \le n$ ) — the elements of the array $a$ .

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

For each test case, output a single integer representing the minimum possible penalty you will receive.

In the first test case, a possible way to split $a$ is $s=[2,4,5]$ and $t=[1,3]$ . The penalty is $p(s)+p(t)=2 + 1 =3$ .

In the second test case, a possible way to split $a$ is $s=[8,3,1]$ and $t=[2,1,7,4,3]$ . The penalty is $p(s)+p(t)=0 + 1 =1$ .

In the third test case, a possible way to split $a$ is $s=[\,]$ and $t=[3,3,3,3,3]$ . The penalty is $p(s)+p(t)=0 + 0 =0$ .