Problem:
Score : 400 points

Problem Statement

You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?

Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.

Note

When a positive integer A divides a positive integer B, A is said to a divisor of B.
For example, 6 has four divisors: 1, 2, 3 and 6.

Constraints

1 \leq N \leq 100

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of the Shichi-Go numbers that are divisors of N!.

Sample Input 1

9

Sample Output 1

0

There are no Shichi-Go numbers among the divisors of 9! = 1 \times 2 \times ... \times 9 = 362880.

Sample Input 2

10

Sample Output 2

1

There is one Shichi-Go number among the divisors of 10! = 3628800: 32400.

Sample Input 3

100

Sample Output 3

543

Solution in Python (3.4.3) (judge verdict: Runtime Error)

import math
prime = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]

M = N
fac_lst = []
for i in prime:
    count = 0
    M = N
    while M % i == 0:
        M = M // i
        count += 1
    fac_lst.append(count)
    
count0, count1, count2, count3, count4 = 0, 0, 0, 0,0
for i in fac_lst:
    if i >= 74:
        count0 += 1
        count1 += 1
        count2 += 1
        count3 += 1
        count4 += 1
    elif i >= 24:
        count1 += 1
        count2 += 1
        count3 += 1
        count4 += 1
    elif i >= 14:
        count2 += 1
        count3 += 1
        count4 += 1
    elif i >= 4:
        count3 += 1
        count4 += 1
    elif i >= 2:
        count4 += 1
count_lst = [count0, count1, count2, count3, count4]

ans0 = count_lst[0]
ans1 = count_lst[1] * (count_lst[4] - 1)
ans2 = count_lst[2] * (count_lst[3] - 1)
ans3 = (count_lst[3]*(count_lst[3]-1) // 2) * (count_lst[4] - 2)
if ans1 < 0:
    ans1 = 0
if ans2 < 0:
    ans2 = 0
if ans3 < 0:
    ans3 = 0
print(ans0+ans1+ans2+ans3)