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'
pr = []
pr2 = Array.new(25, 0)
Prime.each(100) do |prime|
  pr.push(prime)
end
nu = gets.to_i
for i in 1..nu
  inst = i
  for j in 0..24
    while inst % pr[j] == 0
      pr2[j] += 1
      inst /= pr[j]
    end
    if inst == 1
      break
    end
  end
end

ans = 0
if pr2[0] >= 74
  ans += 1
end
inst1 = 0
inst2 = 0
inst3 = 0
inst4 = 0
for i in 0..24
  if pr2[i] >= 2
    inst1 += 1
  end
  if pr2[i] >= 24
    inst2 += 1
  end
  if pr2[i] >= 4
    inst3 += 1
  end
  if pr2[i] >= 14
    inst4 += 1
  end
end
ans += inst2 * (inst1 - 1)
ans += inst4 * (inst3 - 1)
ans += ((inst3 * (inst3 - 1)) / 2) * (inst1 - 2)
puts ans