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())
e = [0] * (N+1)
for i in range(2, N+1):
    cur = i
    for j in range(2, i+1):
        while cur % j == 0:
            e[j] += 1
            cur //= j
def num(m): # e の要素のうち m-1 以上のものの個数
    return len(list(filter(lambda x: x >= m-1, e)))

print(num(75) + num(25) * (num(3) - 1) + num(15) * (num(5) - 1)+ num(5) * (num(5) - 1) * (num(3) - 2) // 2)