Given a sequence of distinct numbers **a _{1}**,

**a**, ….

_{2}**a**, an inversion occurs if there are indices

_{n}**i < j**such that

**a**.

_{i}> a_{j}For example, in the sequence **2 1 4 3** there are 2 inversions **(2 1)** and **(4 3)**.

The input will be a main sequence of N positive integers. From this sequence, a Derived sequence will be obtained using the following rule. The output is the number of inversions in the derived sequence.

**Rule for forming the derived sequence:**

The derived sequence is formed by the sum of digits in the **base 6** representation of the corresponding number in the input sequence.

Thus for the number **1409**, the base 6 representation is **10305**, and the sum of digits is 1+0+3+0+5 = **9**. The sum of digits may be done in the decimal system, and does not need to be in base 6.

**Boundary Condition(s):**

2 <= N <= 100

1 <= Each integer value <= 10^7

**Input Format:**

The first line contains N.

The second line contains N integers separated by a comma.

**Output Format:**

The first line contains the number of inversions in the derived sequence formed from the main sequence.

**Example Input/Output 1:**

Input:

5

62,85,44,21,36

Output:

8

Explanation:

The given sequence is 62,85,44,21,36.

The base 6 representations of the integers in the above sequence are **142, 221, 112, 33** and **100**.

The derived sequence is **7, 5, 4, 6 **and** 1** (corresponding to the sum of digits in the base 6 representation).

The number of inversions in the above derived sequence is **8** and they are given below.

(7, 5), (7, 4), (7, 6), (7, 1), (5, 4), (5, 1), (4, 1) and (6, 1).

**Example Input/Output 2:**

Input:

8

39,46,999,22,19,136,91,81

Output:

11

Explanation:

The given sequence is 39,46,999,22,19,136,91,81.

The base 6 representations of the integers in the above sequence are **103, 114, 4343, 34, 31, 344, 231** and **213**.

The derived sequence is **4, 6, 14, 7, 4, 11, 6 **and** 6** (corresponding to the sum of digits in the base 6 representation).

The number of inversions in the above derived sequence is **11** and they are given below.

(6, 4), (14, 7), (14, 4), (14, 11), (14, 6), (14, 6), (7, 4), (7, 6), (7, 6), (11, 6) and (11, 6).

#include<stdio.h> #include <stdlib.h> int main() { int n, k, s = 0; scanf("%d", & n); int a[n]; for (int i = 0; i < n; i++) { scanf("%d,", & k); while (k > 0) { s += k % 6; k /= 6; } a[i] = s; s = 0; } for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) s += a[i] > a[j]; } printf("%d ", s); }

```
def f(k):
return 0 if k==0 else k%6+f(k//6)
n,s=int(input()),0
a=[f(int(i)) for i in input().split(',')]
for i in range(len(a)):
for j in a[i:]:
s+=a[i]>j
print(s)
```