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 PyPy3 (7.3.0) (judge verdict: Accepted)

def p_factorization(n):
    if n == 1: return {}
    pf_cnt = {}
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i == 0:
            cnt = 0
            while temp%i == 0:
                cnt += 1
                temp //= i
            pf_cnt[i] = cnt

    if temp != 1: pf_cnt[temp] = 1
    return pf_cnt


n = int(input())
pd = {}
for i in range(1,n+1):
    curr_p = p_factorization(i)
    for p,cnt in curr_p.items():
        pd.setdefault(p,0)
        pd[p] += cnt

ans = 0

cnt3 = 0
cnt5 = 0
cnt15 = 0
cnt25 = 0
cnt75 = 0

for v in pd.values():
    if v >= 74: cnt75 += 1
    if v >= 24: cnt25 += 1
    if v >= 14: cnt15 += 1
    if v >= 4: cnt5 += 1
    if v >= 2: cnt3 += 1


ans += cnt75
ans += max(cnt15*(cnt5-1), 0)
ans += max(0,cnt25*(cnt3-1))
ans += max(0,(cnt3-2)*(cnt5)*(cnt5-1)//2)
print(ans)