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 Python (3.4.3) (judge verdict: Accepted)

import math
N = int(input())

dic = dict() #素因数分解してつっこむリスト
def trial_division(n):
    tmp = int(math.sqrt(n)) + 1
    a = n
    for num in range(2, tmp):
        while n%num == 0: #割り切れる時
            if not num in dic:
                dic[num] = 1
            else:
                dic[num] += 1
            n //= num
    if n != 1:
        if not n in dic:
            dic[n] = 1
        else:
            dic[n] += 1

    return dic

# trial_division(5)
for i in range(2, N+1):
    trial_division(i)
    # print (i, dic)
    
# print (dic)
count = 0
for i in dic:
    if dic[i] >= 74:
        count += 1
    if dic[i] >= 24:
        for j in dic:
            if i != j and dic[j] >= 2:
                count += 1
    if dic[i] >= 14:
        for j in dic:
            if i != j and dic[j] >= 4:
                count += 1
    if dic[i] >= 4:
        for j in dic:
            if i < j and dic[j] >= 4:
                for k in dic:
                    if i != k and j != k and dic[k] >= 2:
                        count += 1

print (count)