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

hash = Hash.new(0)
(2..N).each do |i|
    Prime.prime_division(i).each do |pr, n|
        hash[pr] += n
    end
end
nums = hash.values.select{|n| n>=2 }
len = nums.length

divs = [1,3,5,15,25,75]

dp = Array.new(len+1){Array.new(76, 0)}
dp[0][1] = 1
len.times do |i|
    divs.each do |d1|
        divs.each do |d2|
            next if d1*d2 > 75
            if nums[i]+1 >= d2
                dp[i+1][d1*d2] += dp[i][d1]
            end
        end
    end
end

puts dp[len][75]