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'

N = gets.to_i

# N!の素因数分解を行う
h = Hash.new(0)
1.upto(N) do |i|
  (i.prime_division).each do |k|
    h[k[0]] += k[1]
  end
end
# p h

cnt = Hash.new(0)
h.each do |k, v|
  [2, 4, 14, 24, 74].each do |t| 
    cnt[t]+=1 if v >= t
  end
end

# @satanic0258さんのツイートを見て、これか～ってなった。丸写しかな。
ans = 0
ans += cnt[4] * ( cnt[4] - 1 ) / 2 * ( cnt[2] - 2 )
ans += cnt[14] * ( cnt[4] - 1 )
ans += cnt[24] * ( cnt[2] - 1 )
ans += cnt[74]

p ans