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)

import math
import collections

def prime_decomposition(n):
    i = 2
    table = []
    while i * i <= n:
        while n % i == 0:
            n /= i
            table.append(i)
        i += 1
    if n > 1:
        n = int(n)
        table.append(n)
    return table

def num(y):
    count = 0
    for i in x:
        if x[i] >= y:
            count += 1
    return count

N = int(input())
li = []
for i in range(1, N+1):
    li += prime_decomposition(i)
# print(collections.Counter(li))
x = collections.Counter(li)
ans = num(74) + num(24) * (num(2) - 1) + num(14) * (num(4) - 1)  + num(4) * (num(4) - 1) * (num(2) - 2) // 2
print(ans)