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
cnt = Array.new(50,0)
for i in 1..n
    for j in 2..49
        while i%j == 0
            i /= j
            cnt[j] += 1
        end
    end
end
a = Array.new(5,0)
cnt.each do |num|
    a[0] += 1 if num >= 2
    a[1] += 1 if num >= 4
    a[2] += 1 if num >= 14
    a[3] += 1 if num >= 24
    a[4] += 1 if num >= 74
end
ans = 0
ans += a[1]*(a[1]-1)/2*(a[0]-2) if a[0] >= 3 && a[1] >= 2
ans += a[2]*(a[1]-1) if a[2] >= 1 && a[1] >= 2
ans += a[3]*(a[0]-1) if a[3] >= 1 && a[0] >= 2
ans += a[4]
puts ans