human
In fact, the problems E1 and E2 do not have much in common. You should probably think of them as two separate problems.

You are given an integer array a[1 … n] = [a_1, a_2, …, a_n].

Let us consider an empty [deque](https://tinyurl.com/pfeucbux) (double-ended queue). A deque is a data structure that supports adding elements to both the beginning and the end. So, if there are elements [3, 4, 4] currently in the deque, adding an element 1 to the beginning will produce the sequence [\color{red}{1}, 3, 4, 4], and adding the same element to the end will produce [3, 4, 4, \color{red}{1}].

The elements of the array are sequentially added to the initially empty deque, starting with a_1 and finishing with a_n. Before adding each element to the deque, you may choose whether to add it to the beginning or to the end.

For example, if we consider an array a = [3, 7, 5, 5], one of the possible sequences of actions looks like this:    1. |  add 3 to the beginning of the deque: |  deque has a sequence [\color{red}{3}] in it.
  
---|---|---  
  2. |  add 7 to the end of the deque: |  deque has a sequence [3, \color{red}{7}] in it.
  
  3. |  add 5 to the end of the deque: |  deque has a sequence [3, 7, \color{red}{5}] in it.
  
  4. |  add 5 to the beginning of the deque: |  deque has a sequence [\color{red}{5}, 3, 7, 5] in it.
  
  
Find the minimal possible number of inversions in the deque after the whole array is processed. 

An inversion in sequence d is a pair of indices (i, j) such that i < j and d_i > d_j. For example, the array d = [5, 3, 7, 5] has exactly two inversions — (1, 2) and (3, 4), since d_1 = 5 > 3 = d_2 and d_3 = 7 > 5 = d_4.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases.

The next 2t lines contain descriptions of the test cases. 

The first line of each test case description contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — array size. The second line of the description contains n space-separated integers a_i (-10^9 ≤ a_i ≤ 10^9) — elements of the array.

It is...
Solve the task in PYTHON3.
 ... Give me a Python code to find the minimal possible number of inversions in the deque after the whole array is processed?
assistant
#Here is Python code to find the minimal possible number of inversions in the deque after the whole array is processed.

try:
        import sys
        from functools import lru_cache, cmp_to_key
        from heapq import merge, heapify, heappop, heappush
        # from math import *
        from collections import defaultdict as dd, deque, Counter as C
        from itertools import combinations as comb, permutations as perm
        from bisect import bisect_left as bl, bisect_right as br, bisect, insort
        from time import perf_counter
        from fractions import Fraction
        import copy
        from copy import deepcopy
        import time
        starttime = time.time()
        mod = int(pow(10, 9) + 7)
        mod2 = 998244353

        def data(): return sys.stdin.readline().strip()
        def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
        def L(): return list(sp())
        def sl(): return list(ssp())
        def sp(): return map(int, data().split())
        def ssp(): return map(str, data().split())
        def l1d(n, val=0): return [val for i in range(n)]
        def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
        def A2(n,m): return [[0]*m for i in range(n)]
        def A(n):return [0]*n
        # input = stdin.buffer.readline
        # I = lambda : list(map(int,input().split()))
        # import sys
        # input=sys.stdin.readline
        

        sys.stdin = open("input.txt", "r")
        sys.stdout = open("output.txt", "w")
except:
        pass

for _ in range(L()[0]):
        n = L()[0]
        A = L()

        rank = {}
        for i,v in enumerate(sorted(A)):
                rank[v] = i + 1 
         
        A = [rank[ele] for ele in A]
        BIT = [0 for i in range(n+10)]
        def update(i):
                while i <= n:
                        BIT[i] += 1
                        i += i&(-i)
        def get_sum(i):
                s = 0
                while(i):
                        s += BIT[i]
                        i -= i&(-i)
                return s

        q = deque()
        q.append(A[0])
        update(A[0])
        cnt = 1
       ...