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: Wrong Answer)

def solve(n)
  a= Array.new(n+1, 1)
  for i in 2..n
    if a[i] == 1
      for j in ((i*2)..n).step(i)
        a[j] = 0
      end
    end
  end
  prime = []
  a.each.with_index do |v, index|
    prime.push(index) if v != 0 && index >= 2
  end

  b= Array.new(n+1, 0)
  for i in 2..n
    if a[i] == 1
      a[i] = 0
      j = i
      while j < n
        b[i] += n / j
        j *= i
      end
    end
  end
  
  count = 0
  for i in 0..n
    for j in (i+1)..n
      for k in 0..n
        next if i == k || j == k
        count += 1 if b[i] >= 4 && b[j] >= 4 && b[k] >= 2
      end
      count += 1 if b[i] >= 14 && b[j] >= 4
      count += 1 if b[i] >= 24 && b[j] >= 2
    end
    count += 1 if b[i] >= 74
  end
  
  return count
end


puts solve(gets.to_i)