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'
N=gets.to_i
h = Hash.new(0)
1.upto(N) do |i|
    i.prime_division.each do |a,b|
        h[a] += b
    end
end

nn = 1
1.upto(N) do |i|
    nn *= i
end
ans = []
h.keys.permutation(3) do |a,b,c|
    tmp = a**4 * b**4 * c**2
    if nn % tmp == 0
        ans << tmp  
    end
end
h.keys.permutation(2) do |a,b|
    tmp = a**24 * b**2
    if nn % tmp == 0
        ans << tmp  
    end
    tmp = a**14 * b**4
    if nn % tmp == 0
        ans << tmp  
    end
end


h.each do |k,v|
    ans << k ** 74 if v >= 74
end

p ans.uniq.size