human
It is the easy version of the problem. The only difference is that in this version n = 1.

In the cinema seats can be represented as the table with n rows and m columns. The rows are numbered with integers from 1 to n. The seats in each row are numbered with consecutive integers from left to right: in the k-th row from m (k - 1) + 1 to m k for all rows 1 ≤ k ≤ n.

1| 2| ⋅⋅⋅| m - 1| m  
---|---|---|---|---  
m + 1| m + 2| ⋅⋅⋅| 2 m - 1| 2 m  
2m + 1| 2m + 2| ⋅⋅⋅| 3 m - 1| 3 m  
\vdots| \vdots| \ddots| \vdots| \vdots  
m (n - 1) + 1| m (n - 1) + 2| ⋅⋅⋅| n m - 1| n m  
The table with seats indices

There are nm people who want to go to the cinema to watch a new film. They are numbered with integers from 1 to nm. You should give exactly one seat to each person.

It is known, that in this cinema as lower seat index you have as better you can see everything happening on the screen. i-th person has the level of sight a_i. Let's define s_i as the seat index, that will be given to i-th person. You want to give better places for people with lower sight levels, so for any two people i, j such that a_i < a_j it should be satisfied that s_i < s_j.

After you will give seats to all people they will start coming to their seats. In the order from 1 to nm, each person will enter the hall and sit in their seat. To get to their place, the person will go to their seat's row and start moving from the first seat in this row to theirs from left to right. While moving some places will be free, some will be occupied with people already seated. The inconvenience of the person is equal to the number of occupied seats he or she will go through.

Let's consider an example: m = 5, the person has the seat 4 in the first row, the seats 1, 3, 5 in the first row are already occupied, the seats 2 and 4 are free. The inconvenience of this person will be 2, because he will go through occupied seats 1 and 3.

Find the minimal total inconvenience (the sum of inconveniences of all people), that is possible to have by giving places for all people...
The above is tricky. Write me a correct solution in PYTHON3.
 What is a Python program to find the minimal total inconvenience (the sum of inconveniences of all people), that is possible to have by giving places for all people?
assistant
#Here is Python program to find the minimal total inconvenience (the sum of inconveniences of all people), that is possible to have by giving places for all people.

#!/usr/bin/env python3
import sys
import getpass  # not available on codechef
import math, random
import functools, itertools, collections, heapq, bisect
from collections import Counter, defaultdict, deque
input = sys.stdin.readline  # to read input quickly

# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy

M9 = 10**9 + 7  # 998244353
yes, no = "YES", "NO"
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)]  # hexagonal layout
MAXINT = sys.maxsize

# if testing locally, print to terminal with a different color
OFFLINE_TEST = getpass.getuser() == "hkmac"
# OFFLINE_TEST = False  # codechef does not allow getpass
def log(*args):
    if OFFLINE_TEST:
        print('\033[36m', *args, '\033[0m', file=sys.stderr)

def solve(*args):
    # screen input
    if OFFLINE_TEST:
        log("----- solving ------")
        log(*args)
        log("----- ------- ------")
    return solve_(*args)

def read_matrix(rows):
    return [list(map(int,input().split())) for _ in range(rows)]

def read_strings(rows):
    return [input().strip() for _ in range(rows)]

def minus_one(arr):
    return [x-1 for x in arr]

def minus_one_matrix(mrr):
    return [[x-1 for x in row] for row in mrr]

# ---------------------------- template ends here ----------------------------


class FenwickTree:
    # also known as Binary Indexed Tree
    # binarysearch.com/problems/Virtual-Array
    # https://leetcode.com/problems/create-sorted-array-through-instructions
    # may need to be implemented again to reduce constant factor
    def __init__(self, bits=31):
        self.c = defaultdict(int)
        self.LARGE = 2**bits

    def update(self, x, increment):
        x += 1  # to avoid infinite loop at x > 0
        while x <= self.LARGE:
            # increase by the greatest power of two that divides x
            self.c[x] += increment
            x += x & -x

    def query(self, x):
        x += 1  #...