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())

primenum = [0]*101

def make_divisor(n):
    res = []
    for i in range(1,int(n**0.5+1)):
        if i*i == n:
            res.append(i)
        else:
            res.append(i)
            res.append(n//i)
    return res

def fact(n):
    if n==1:
        return 
    else:
        divs = make_divisor(n)
        divs.sort()
        tmp = n
        idx = 1
        while(tmp>1):
            while (tmp%divs[idx]==0):
                primenum[divs[idx]] += 1
                tmp //= divs[idx]
            idx += 1
        fact(n-1)
        return 


fact(N)

ans = 0
primenum.sort(reverse = True)
idx = 0
while primenum[idx]>=74: idx += 1
ans += idx

idx2 = 0
while primenum[idx]>=24: idx += 1
while primenum[idx+idx2]>=2:idx2 += 1
ans += idx*(idx-1+idx2)

idx = 0
idx2 = 0
while primenum[idx]>=14: idx += 1
while primenum[idx+idx2]>=4:idx2 += 1
ans += idx*(idx-1+idx2)

idx = idx+idx2
idx2 = 0
while primenum[idx+idx2]>=2: idx2 += 1
ans += idx*(idx-1)//2*(idx-2+idx2)
print(ans)