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())
p = {}
for i in range(2, n + 1):
    j = i
    for d in range(2, j + 1):
        while j % d == 0:
            p[d] = p.get(d, 0) + 1
            j //= d

def doit(ds, ps):
    if ds == []:
        return 1
    if ps == []:
        return 0
    ret = doit(ds, ps[1:])
    if ps[0] + 1 >= ds[0]:
        ret += doit(ds[1:], ps[1:])
    return ret

ret = 0
for ds in ([3, 5, 5], [5, 3, 5], [5, 5, 3], [5, 15], [15, 5], [3, 25], [25, 3], [75]):
    ret += doit(ds, list(p.values()))
print(ret)
