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 (2.7.6) (judge verdict: Accepted)

N = input()
C = [0]*101
for i in range(1, N+1):
  f = 2
  while i>1:
    if i%f==0:
      C[f] += 1
      i /= f
    else:
      f += 1
ans = 0
for i in range(101):
  if C[i]>=2:
    for j in range(101):
      if j != i and C[j]>=4:
        for k in range(j):
          if k != i and C[k]>=4:
            ans += 1
for i in range(101):
  if C[i]>=2:
    for j in range(101):
      if j != i and C[j]>=24:
        ans += 1
for i in range(101):
  if C[i]>=14:
    for j in range(101):
      if j != i and C[j]>=4:
        ans += 1
for i in range(101):
  if C[i]>=74:
    ans += 1
print ans