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 nprf(n):
    pf = {}
    for i in range(2, n+1):
        for j in range(2, i+1):
            while i % j == 0:
                if not j in pf:
                    pf[j] = 0
                pf[j] += 1
                i //= j
    return pf

pf = nprf(n)

def f(n):
    count = 0
    for v in pf.values():
        if v >= n-1:
            count += 1
    return count

f1 = f(75)
f2 = f(25) * (f(3)-1)
f3 = f(15) * (f(5)-1)
f4 = f(5) * (f(5)-1) //2 * (f(3)-2)

print(f1 + f2 + f3 + f4)