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 (2.7.1) (judge verdict: Wrong Answer)

require "prime"
n = gets.chomp.to_i

prime_factors = n.times.inject({}) do |res, i|
  res.merge(Prime.prime_division(i + 1).to_h) do |key, v0, v1|
    v0 + v1
  end
end 

ge_4 = prime_factors.select {|_k, v| v >= 4}.size
ge_2 = prime_factors.select {|_k, v| v >= 2}.size

if ge_4 < 2
  puts 0
else
  puts ge_4 * (ge_4 - 1) / 2 * (ge_2 - 2)
end