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)

from collections import defaultdict, deque, Counter
import sys
input = sys.stdin.readline


def getPrimes(n):
    isPrime = [True] * n
    ret = []
    for p in range(2, n):
        if not isPrime[p]:
            continue
        ret.append(p)
        for i in range(p, n, p):
            isPrime[i] = False
    return ret


def main():
    N = int(input())
    P = getPrimes(N)
    C = Counter()

    for i in range(1, N + 1):
        cnt = Counter()
        for p in P:
            while i % p == 0:
                i //= p
                cnt[p] += 1
        C += cnt

    d = {3: 0, 5: 0, 15: 0, 25: 0, 75: 0}
    for k in d.keys():
        for v in C.values():
            if v >= k - 1:
                d[k] += 1

    ans = 0
    ans += (d[5] * (d[5] - 1) // 2) * (d[3] - 2)
    ans += d[15] * (d[5] - 1)
    ans += d[25] * (d[3] - 1)
    ans += d[75]
    print(ans)


if __name__ == '__main__':
    main()