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 (2.4.0) (judge verdict: Accepted)

n = int(input())
prits = [0 for i in range(n+1)]

for i in range(1,n+1):
    k = 2
    while k*k<=i:
        while i%k == 0:
            i = i//k
            prits[k] += 1
        k += 1
    if i>1 :
        prits[i] += 1
#print(prits)

a,b,c,d,e = [],[],[],[],[]

for i in range(len(prits)):
    num = prits[i]
    if num >=2:
        e.append(i)
    if num >=4:
        d.append(i)
    if num >=14:
        c.append(i)
    if num >=24:
        b.append(i)
    if num >=74:
        a.append(i)

#print(len(a),len(b),len(c),len(d),len(e))

ans = 0
ans += len(a)
ans += len(c)*(len(d)-1)
ans += len(b)*(len(e)-1)
ans += len(d)*(len(d)-1)*(len(e)-2)//2

print(ans)