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())
l_prm = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
l_prm = [i for i in l_prm if i<=n/2]
l_new = []
for i in l_prm:
    a = n
    counter = 0
    b = a
    c=1
    while b!=0:
        b = a//i**c
        counter += b
        c += 1
    l_new.append(counter)
ans=0
def num(j,lst):
    return(len([i for i in lst if i>=j]))
ans += num(4,l_new) * (num(4,l_new)-1)/2 *(num(2,l_new)-2)
ans += num(14,l_new)*(num(4,l_new)-1)
ans += num(24,l_new)*(num(2,l_new)-1)
ans += num(74,l_new)
print(int(ans))