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)

from collections import Counter

n = int(input())

def prime_factorize(n):
    a = []
    while n % 2 == 0:
        a.append(2)
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            a.append(f)
            n //= f
        else:
            f += 2
    if n != 1:
        a.append(n)
    return a

li = []

for i in range(1,n+1):
    li += prime_factorize(i)

C = Counter(li)

a,b,c,d,e = 0,0,0,0,0

for i in C.values():
    if i >= 74:
        e += 1
    elif i >= 24:
        d += 1
    elif i >= 14:
        c += 1
    elif i >= 4:
        b += 1
    elif i >= 2:
        a += 1

point = 0
point += e
point += (d+e)*(a+b+c) + (d+e)*(d+e-1)
point += (c+d+e)*b + (c+d+e)*(c+d+e-1)
point += a*((b+c+d+e)*(b+c+d+e-1)//2) + (b+c+d+e)*((b+c+d+e-1)*(b+c+d+e-2)//2)

print(point)