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 Python (3.4.3) (judge verdict: Accepted)

N = int(input())
primes = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47]

two = sum(p<=N//2 for p in primes) # 2
four = sum(p<=N//4 for p in primes) if N>=12 else sum([N>=9, N>=6]) # 4
ft = sum([N>=91, N>=60, N>=30, N>=16]) # 14
tf = sum([N>=100, N>=54, N>=28]) # 24
sf = N>=78 # 74

tff = four * (four-1) * (two-2) // 2 # 2,4,4
ttf = tf * (two-1) # 2,24
fft = ft * (four-1) # 4,14
sfo = sf # 74

print(tff+ttf+fft+sfo)