101143: [AtCoder]ABC114 D - 756
Memory Limit:256 MB
Time Limit:2 S
Judge Style:Text Compare
Creator:
Submit:0
Solved:0
Description
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 $N!$.
Sample Input 1
9
Sample Output 1
0
There are no Shichi-Go numbers among the divisors of $9! = 1 \times 2 \times ... \times 9 = 362880$.
Sample Input 2
10
Sample Output 2
1
There is one Shichi-Go number among the divisors of $10! = 3628800$: $32400$.
Sample Input 3
100
Sample Output 3
543