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(m):
    pf = {}
    for i in range(2, m + 1):
        for j in range(2, i + 1):
            while i % j == 0:
                pf[j] = pf.get(j, 0) + 1
                i //= j
    return pf

def c(P, m):
    count = 0
    for i in P:
        if P[i] >= m - 1:
            count += 1
    return count   
    
P = nprf(n)

f75 = c(P, 75)
f253 = c(P, 25) * (c(P, 3) - 1)
f155 = c(P, 15) * (c(P, 5) - 1)
f553 = int(c(P, 5) * (c(P, 5) - 1) * (c(P, 3) - 2) * 0.5)

print(f75 + f253 + f155 + f553)