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())
f=list()

for i in range(1,n+1):
  b=2
  while b*b <= i:
    while i%b==0:
      i //= b
      f.append(b)
    b += 1
  if i > 1:
    f.append(i)

o2=o3=o4=o14=o24=o74=0

for i in range(2,n+1):
  d=f.count(i)
  if d>=2:
    o2 += 1
  if d>=4:
    o4 += 1
  if d>=14:
    o14 += 1
  if d>=24:
    o24 += 1
  if d>=74:
    o74 += 1

print(o4*(o4-1)//2*(o2-2)+o14*(o4-1)+o24*(o2-1)+o74)