CF1919C.Grouping Increases

传统题 时间 2000 ms 内存 256 MiB 7 尝试 3 已通过 1 标签

Grouping Increases

You are given an array aa of size nn. You will do the following process to calculate your penalty:

  1. Split array aa into two (possibly empty) subsequences^\dagger ss and tt such that every element of aa is either in ss or tt^\ddagger.
  2. For an array bb of size mm, define the penalty p(b)p(b) of an array bb as the number of indices ii between 11 and m1m - 1 where bi<bi+1b_i \lt b_{i + 1}.
  3. The total penalty you will receive is p(s)+p(t)p(s) + p(t).

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

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

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

Input

Each test contains multiple test cases. The first line contains a single integer tt (1t1041 \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 nn (1n21051\le n\le 2\cdot 10^5) — the size of the array aa.

The second line contains nn integers a1,a2,,ana_1, a_2, \ldots, a_n (1ain1 \le a_i \le n) — the elements of the array aa.

It is guaranteed that the sum of nn over all test cases does not exceed 21052\cdot 10^5.

Output

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

Note

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

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

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

Samples

5
5
1 2 3 4 5
8
8 2 3 1 1 7 4 3
5
3 3 3 3 3
1
1
2
2 1
3
1
0
0
0

在线编程 IDE

建议全屏模式获得最佳体验