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(raw_input())
prime_in_N = {}
for n in range(2, N+1):
  while True:
    bn = n
    for p in (2, 3, 5, 7):
      if n % p == 0:
        n /= p
        if not p in list(prime_in_N.keys()):
          prime_in_N[p] = 0
        prime_in_N[p] += 1
    if bn == n:
      break
  if n > 7:
    if not n in list(prime_in_N.keys()):
      prime_in_N[n] = 0
    prime_in_N[n] += 1
ans = 0
for a in list(prime_in_N.keys()):
  if prime_in_N[a] >= 74:
	ans += 1
for a in list(prime_in_N.keys()):
  if prime_in_N[a] >= 2:
    for b in set(prime_in_N.keys())-set([a]):
      if prime_in_N[b] >= 24:
        ans += 1
for a in list(prime_in_N.keys()):
  if prime_in_N[a] >= 4:
    for b in set(prime_in_N.keys())-set([a]):
      if prime_in_N[b] >= 14:
        ans += 1
tmp_ans = 0
for a in list(prime_in_N.keys()):
  if prime_in_N[a] >= 2:
    for b in set(prime_in_N.keys())-set([a]):
      if prime_in_N[b] >= 4:
        for c in set(prime_in_N.keys())-set([a])-set([b]):
      		if prime_in_N[c] >= 4:
        		tmp_ans += 1
print ans+tmp_ans/2