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'

#組み合わせ（小さい数用）
def combi(n, m)
  return 1 if m == 0
  return n if m == 1
  return combi(n, n - m) if m > n - m
  fact_s(n, n - m + 1) / fact_s(m, 1)
end
#階乗（小さい数用）
def fact_s(n, m)
  (m..n).to_a.inject(&:*)
end

N = gets.to_i

pr = Hash.new{|h, k| h[k] = 0}
N.downto(2){|n|
  n.prime_division.each{|k, v|
    pr[k] += v
  }
}
ans = 0
cnt3 = pr.count{|k, v| v >= 2}
cnt5 = pr.count{|k, v| v >= 4}
cnt15= pr.count{|k, v| v >= 14}
cnt25= pr.count{|k, v| v >= 24}
cnt75= pr.count{|k, v| v >= 74}
ans += combi(cnt5, 2) * combi(cnt3-2, 1) if cnt5 > 1
ans += combi(cnt15, 1) * combi(cnt5-1, 1) if cnt15 > 0
ans += combi(cnt25, 1) * combi(cnt3-1, 1) if cnt25 > 0
ans += combi(cnt75, 1) if cnt75 > 0

puts ans