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: Accepted)

n = int(input())

from collections import defaultdict
from math import sqrt

A = defaultdict(int)

def factorize(n):    
    i = 2
    
    while n % i == 0:
        n //= i
        A[i] += 1
    i += 1

    while n != 1 and i <= sqrt(n):
        while n % i == 0:
            n //= i
            A[i] += 1
        i += 2
    if n != 1:
        A[n] += 1

    return A

for i in range(1, n+1):
    factorize(i)

ans = 0

c_74 = 0
c_2 = 0
c_24 = 0
c_4 = 0
c_14 = 0

for k, v in A.items():
    if v >= 74:
        c_74 += 1
    if v >= 2:
        c_2 += 1
    if v >= 24:
        c_24 += 1
    if v >= 4:
        c_4 += 1
    if v >= 14:
        c_14 += 1

ans += c_74
ans += max(0, (c_2 - 1) * c_24)
ans += max(0, (c_2 - 2) * c_4 * (c_4 - 1) // 2)
ans += max(0, c_14 * (c_4 - 1))

print(ans)