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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

import Data.List

main = readLn >>= print.solve

solve n =
    let xs = map length$group$factorize p100$product [1..n]
        [a,b,c,d,e] = map (over xs) [2,4,14,24,74]
    in div (b * (b-1)) 2 * (a-2) + c * (b-1) + d * (a-1) + e

over xs n = length(filter(>=n)xs)

p100 = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47]

factorize [] _ = []
factorize pr@(p:ps) n = if n`rem`p==0 then p:(factorize pr (n`quot`p)) else factorize ps n