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
import sys

def get_pfn(N):
  e = [0] * (N+1)
  for i in range(2, N+1):
    cur = i
    for j in range(2, i+1):
      while cur % j == 0:
        e[j] += 1
        cur //= j
  return e


def num(n, e):
  return len(list(filter(lambda x: x >= n-1, e)))

def main():
  input = sys.stdin.readline
  N = int(input())
  e = get_pfn(N)
  print(num(75, e) +\
        num(25, e) * (num(3, e) - 1) +\
        num(15, e) * (num(5, e) - 1) +\
        num(5, e) * (num(5, e) - 1) * (num(3, e) - 2) // 2)

  
if __name__ == '__main__':
  main()