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"
#input of int(split by space)
def get_i()
  return gets.chomp.split(" ").map(&:to_i)
end
#input of float(split by space)
def get_f()
  return gets.chomp.split(" ").map(&:to_f)
end
#input of string(split by space)
def get()
  return gets.chomp.split(" ")
end
#input of string(split per one character)
def get_nsp()
  return gets.chomp.split("")
end
#yes or no decision
def yn_judge(bool,y="Yes",n="No")
  return bool ? y : n 
end
#create of array
def array(size1,init=nil,size2=-1)
  if size2==-1
    return Array.new(size1){init}
  else
    return Array.new(size2){Array.new(size1){init}}
  end
end

def max(a,b)
    return a>b ? a : b
end

def min(a,b)
    return a>b ? b : a
end

INF=Float::INFINITY

N=gets.to_i
kind=Hash.new(0)
2.upto(N) do|x|
  primes=Prime.prime_division(x)
  primes.each do|k,v|
    kind[k]+=v
  end
end
p3=0
p5=0
p15=0
p25=0
p75=0
kind.each do|k,v|
  p3+=1 if v>=2
  p5+=1 if v>=4
  p15+=1 if v>=14
  p25+=1 if v>=24
  p75+=1 if v>=74
end
puts p5*(p5-1)/2*(p3-2)+p25*(p3-1)+p75+p15*(p5-1)