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())
D = [0]*(n+1)

def check_div_n(div_list, n):
  return len([d for d in div_list if d>=n and d!=0])

for i in range(2, n+1):
  check_n = i
  for j in range(2, n+1):
    while check_n%j==0:
      D[j] += 1
      check_n = check_n//j
      
ans = check_div_n(D, 74) + \
      (check_div_n(D, 4)-1)*check_div_n(D, 14) + \
      (check_div_n(D, 2)-1)*check_div_n(D, 24) + \
      check_div_n(D, 4)*(check_div_n(D, 4)-1)//2 * \
      (check_div_n(D, 2)-2)
  
print(ans)