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

factoring = Array.new(n + 1, 0)

2.upto(n) do |i|
  2.upto(i) do |j|
    while i % j == 0
      factoring[j] += 1
      i /= j
    end
  end
end

cnt = 0

n.times do |i|
  next if factoring[i] < 4
  i.times do |j|
    next if factoring[j] < 4
    n.times do |k|
      cnt += 1 if k != i && k != j && factoring[k] > 1
    end
  end
end

n.times do |i|
  next if factoring[i] < 24
  n.times do |j|
    cnt += 1 if i != j && factoring[j] > 1
  end
end

n.times do |i|
  next if factoring[i] < 14
  n.times do |j|
    cnt += 1 if i != j && factoring[j] > 3
  end
end

n.times do |i|
  cnt += 1 if factoring[i] > 73
end

puts cnt