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 Ruby (2.3.3) (judge verdict: Accepted)

require 'prime'

def factorial(n)
  return 1 if n == 0
  (1..n).inject(1,:*)
end

def nCk(n, k)
  factorial(n) / (factorial(k) * factorial(n - k))
end

n = gets.to_i
factors = factorial(n).prime_division
ans = 0
n1 = factors.size
n3, n5, n15, n25, n75 = 0, 0, 0, 0, 0
factors.each do |f|
  n3 += 1 if f[1] >= 2
  n5 += 1 if f[1] >= 4
  n15 += 1 if f[1] >= 14
  n25 += 1 if f[1] >= 24
  n75 += 1 if f[1] >= 74
end
ans += nCk(n1, (n1 - 75)) if n1 >= 75
ans += (n3 - n5) * ((n5 * (n5 - 1)) / 2) + ((n5 * (n5 - 1) * (n5 - 2)) / 6) * 3 if n3 >= 3 && n5 >= 2
ans += (n5 - n15) * n15 + (n15 * (n15 - 1)) if n5 >= 2 && n15 >= 1
ans += (n3 - n25) * n25 + (n25 * (n25 - 1)) if n3 >= 2 && n25 >= 1
ans += n75
puts ans