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'
a = [*1..gets.to_i].inject(Hash.new{0}){|a, b| b.prime_division.each{|k,v| a[k] += v}; a}.values.map(&:succ).select{|a| a >= 3}
ans = 0r
[
  [3,5,5],
  [3,25],
  [5,15],
  [75]
].each do |b|
  a.permutation(b.length).each do |c|
    z = true
    (0...b.length).each do |i|
      if b[i] > c[i]
        z = false
        break
      end
    end
    ans += (b.length == 3 ? 1/2r : 1r) if z
  end
end
p ans.to_i