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
def comb(n, r)
    f = [1]
    (1..n).each do |i|
        f << f[i - 1]*i
    end
    return f[n] / (f[r] * f[n - r])
end

factors = {}
(1..N).each do |n|
    primes = n.prime_division
    primes.each do |k, v|
        if factors.key?(k)
            factors[k] += v
        else
            factors[k] = v
        end
    end
end

ans = 0
candidate2 = 0
candidate4 = 0
candidate14 = 0
candidate24 = 0
candidate74 = 0

factors.each_value do |v|
    candidate2 += 1 if v >= 2    
    candidate4 += 1 if v >= 4    
    candidate14 += 1 if v >= 14    
    candidate24 += 1 if v >= 24    
    candidate74 += 1 if v >= 74    
end

# (74)
ans += candidate74

# (4, 14)
ans += candidate14 * (candidate4 - 1)

# (2, 24)
ans += candidate24 * (candidate2 - 1)

# (2, 4, 4)
ans += comb(candidate4, 2) * (candidate2 - 2) if candidate4 >= 2

puts ans