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)

N = gets.to_i

pf = [0] * (N + 1)
2.upto(N) do |m|
  2.upto(m) do |i|
    next if m == 1
    while m % i == 0
      m /= i
      pf[i] += 1
    end
  end
end

a = pf.count{|k| k >= 74}
b = pf.count{|k| k >= 24}
c = pf.count{|k| k >= 14}
d = pf.count{|k| k >= 4}
e = pf.count{|k| k >= 2}

p a + b * (e - 1) + c * (d - 1) + d * (d - 1) / 2 * (e - 2)