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)

import sys

def I(): return int(sys.stdin.readline().rstrip())

N = I()

prime = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47]

d = {}  # d[p] = Nがpで最大何回割り切れるか
for p in prime:
    r = 0
    for i in range(1,7):
        r += N//(p**i)
    d[p] = r

A = 0  # d[p] >= 74 たるpの個数
B = 0  # d[p] >= 24 たるpの個数
C = 0  # d[p] >= 14 たるpの個数
D = 0  # d[p] >= 4 たるpの個数
E = 0  # d[p] >= 2 たるpの個数

for p in prime:
    if d[p] >= 74:
        A += 1
    if d[p] >= 24:
        B += 1
    if d[p] >= 14:
        C += 1
    if d[p] >= 4:
        D += 1
    if d[p] >= 2:
        E += 1

# 順に、p**74,p**24×q**2,p**14×q**4,p**4×q**4×r**2 の個数を加える

print(A+B*(E-1)+C*(D-1)+((D*(D-1))//2)*(E-2))