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What is a Python program to find the minimal possible number of inversions in the deque after the whole array is processed given the following:
PYTHON3 solution for "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...":
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#Here is Python program 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
       ...