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)

# Your code here!


n = gets.to_i

arr = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47]
arr2 = Array.new(15,0)

i = 1
n.times{
    key = i
    
    s = 0
    15.times{
        
        
        loop{
            if key%arr[s] == 0
                key/=arr[s]
                arr2[s]+=1
                
            else
                break
            end
        }
        
        
       s+=1 
    }
    
    
    i+=1
}



answer = 0

if arr2[0]>=74
    answer+=1
end

count14 = 0
count4 = 0
count24 = 0
count2 = 0

k = 0
15.times{
    
   if arr2[k]>=24
       count24+=1
   end
   
   if arr2[k]>=14
       count14+=1
   end
   
   if arr2[k]>=2 
       count2+=1
   end
   
   if arr2[k]>=4
       count4+=1
   end

    
  k+=1
}



answer+=count14*(count14-1)+count14*(count4-count14)

answer+=count24*(count24-1)+count24*(count2-count24)


answer+=count4*(count4-1)*(count4-2)/2
answer+=count4*(count4-1)*(count2-count4)/2


p answer