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...
Using python3 can you solve the prior task?

Write me a Python program to find the minimal possible number of inversions in the deque after the whole array is processed?
assistant
#Here is Python program to find the minimal possible number of inversions in the deque after the whole array is processed.

import sys
input = sys.stdin.readline
# sys.setrecursionlimit(400000)
def I(): return input().strip()
def II(): return int(input().strip())
def LI(): return [*map(int, input().strip().split())]
import copy, string, math, time, functools, fractions
from heapq import heappush, heappop, heapify
from bisect import bisect_left, bisect_right
from collections import deque, defaultdict, Counter, OrderedDict
from itertools import permutations, chain, combinations, groupby
from operator import itemgetter
from types import GeneratorType  # for recursion
from typing import Iterable, TypeVar, Union  # for sorted set

# for _ in range(II()):
#     n, m, k = LI()
#     inp = []
#     for i in range(n):
#         inp.append(I())
#     cnt = 0
#     for i in range(n):
#         for j in range(m):
#             if inp[i][j] == '*':
#                 cnt += 1
#     vertex = []
#     for i in range(n):
#         for j in range(m):
#             if i == 0 or j == 0 or j == m - 1:
#                 continue
#             if inp[i][j] == '*':
#                 if i < n - 1:
#                     if inp[i - 1][j - 1] == inp[i + 1][j - 1] == inp[i - 1][j + 1] == inp[i + 1][j + 1] == '*':
#                         continue
#                 if inp[i - 1][j - 1] == inp[i - 1][j + 1] == '*':
#                     vertex.append((i, j))
#
#     def height(i, j):
#         ti, tj = i, j
#         while inp[ti][tj] == '*':
#             ti, tj = ti - 1, tj - 1
#             if ti < 0 or tj < 0:
#                 break
#         l = i - ti
#         ti, tj = i, j
#         while inp[ti][tj] == '*':
#             ti, tj = ti - 1, tj + 1
#             if ti < 0 or tj > m - 1:
#                 break
#         r = i - ti
#         return min(l, r)
#
#     flag = True
#     set_ = set(vertex)
#     print(vertex)
#     for vi, vj in vertex:
#         h = height(vi, vj) - 1
#         # print(vi, vj, h)
#         if h < k:
#             flag = False
#         for i in range(1, h + 1):
#             set_.add((vi - i, vj - i))
#            ...