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.

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!.

Examples

Input

9


Output

0


Input

10


Output

1


Input

100


Output

543
N = int(input())


def factorize(n):
    i, factors = 2, []
    while i * i <= n:
        while n % i == 0:
            n //= i
            factors.append(i)
        i += 1
    if n > 1:
        factors.append(n)
    return factors


e = [0] * (N + 1)
for i in range(1, N + 1):
    for s in factorize(i):
        e[s] += 1


def f(n):
    cnt = 0
    for x in e:
        if x >= n - 1:
            cnt += 1
    return cnt


ans = f(75)
ans += f(25) * (f(3) - 1)
ans += f(15) * (f(5) - 1)
ans += f(5) * (f(5) - 1) * (f(3) - 2) // 2
print(ans)