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: Runtime Error)

class Integer
  def d_list
    return [1] if self == 1
    (1..Math.sqrt(self).to_i).select do |i|
      self%i == 0
    end.map {|e| [e, self/e]}.flatten.uniq.sort
  end
end
N=gets.to_i
ps=[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47]
def com(n,k)
  return 0 if n<=0 || k<=0
  return (1..n).inject(:*)/((1..n-k).inject(:*)*(1..k).inject(:*))
end
hs=Hash.new{0}
for i in 1..N
  ps.each do |e|
    a=i
    while true
      break if a%e!=0
      a/=e
      hs[e]+=1
    end
  end
end
ns=Array.new(N,0)
ps.each do |e|
  hs[e].times do |i|
    ns[i+1]+=1
  end
end
puts com(ns[4],2)*com(ns[2]-2,1)+com(ns[14],1)*com(ns[4]-1,1)+com(ns[24],1)*com(ns[2]-1,1)+com(ns[74],1)