Score : $400$ points

Problem Statement

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.

Note

When a positive integer $A$ divides a positive integer $B$, $A$ is said to a divisor of $B$. For example, $6$ has four divisors: $1, 2, 3$ and $6$.

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 $N!$.

9

0

There are no Shichi-Go numbers among the divisors of $9! = 1 \times 2 \times ... \times 9 = 362880$.

10

1

There is one Shichi-Go number among the divisors of $10! = 3628800$: $32400$.

100

543