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)

#
# File    : D2.rb
# Author  : Kazune Takahashi
# Created : 2018-12-2 21:41:17
# Powered by Visual Studio Code
#

require 'prime'

n = gets.to_i

hash = {}

for i in 2...100
  if i.prime?
    hash[i] = 0
  end
end

for i in 1..n
  i.prime_division.each{|prime, cnt|
    hash[prime] += cnt
  }
end

# p hash

ary = Array.new(100){0}

hash.each{|prime, cnt|
  ary[cnt] += 1
}

98.downto(0){|i|
  ary[i] += ary[i + 1]
}

ans = 0
ans += ary[74]
ans += (ary[4] - 1) * ary[14]
ans += (ary[2] - 1) * ary[24]
ans += (ary[2] - 2) * ary[4] * (ary[4] - 1) / 2

puts ans