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())

# N!を素因数分解をしたときの可能な最大値はNであることに注意
# eは0~Nまでの指数をカウントするもの
e = [0]*(N+1)

# 2からNまでそれぞれ順に素因数分解していく
for i in range(2, N+1):
    cur = i
    # 2~iまでそれぞれで割っていく
    for j in range(2, i+1):
        while cur % j == 0:
            e[j] += 1
            # curはjで割った結果になる
            cur //= j

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

print(num(75) + 
      num(25) * (num(3) - 1) + 
      num(15) * (num(5) - 1) + 
      num(5) * (num(5) - 1) * (num(3) - 2) // 2)
