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())
hurui = [1]*(n+1)
hurui[0] = 0
hurui[1] = 0
sq = int(n**(1/2))+3

for i in range(2,sq):
    for j in range(2,n//i+1):
        hurui[j*i] = 0
p = []

for i in range(n+1):
    if hurui[i]:
        p.append(i)

cnt = {pi:0 for pi in p}

for pi in p:
    tmp = n
    while tmp>0:
        tmp //= pi
        cnt[pi] += tmp
      
cnt2 = {2:0, 4:0, 14:0, 24:0,74:0}

for key in cnt:
    if cnt[key] >= 2:
        cnt2[2]+=1
    if cnt[key] >= 4:
        cnt2[4]+=1
    if cnt[key] >= 14:
        cnt2[14]+=1
    if cnt[key] >= 24:
        cnt2[24]+=1
    if cnt[key] >= 74:
        cnt2[74]+=1

ans = 0

ans += (cnt2[2]-1)*cnt2[24]
ans += (cnt2[4]-1)*cnt2[14]
ans += (cnt2[2]-2)*cnt2[4]*(cnt2[4]-1)//2
ans += cnt2[74]
print(ans)