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
from collections import defaultdict
N = int(input())
d = defaultdict(int)
for n in range(1, N+1):
    m = 2
    while n != 1:
        if n % m == 0:
            d[m] += 1
            n //= m
            m = 2
        else:
            m += 1
ans = 0
# count 3, 5, 15, 25, 75
count = [0, 0, 0, 0, 0]
for v in d.values():
    if v >= 74:
        count[4] += 1
    elif v >= 24:
        count[3] += 1
    elif v >= 14:
        count[2] += 1
    elif v >= 4:
        count[1] += 1
    elif v >= 2:
        count[0] += 1
ans += count[4]
ans += sum(count[3:])*(sum(count)-1)
ans += sum(count[2:])*(sum(count[1:])-1)
ans += sum(count[1:])*(sum(count[1:])-1)//2*(sum(count)-2)
print(ans)