Problem:
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

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

n = gets.chomp.to_i
ans = 0

@ary357 = ['3','5','7']

def retarys(int, ary)
  retary = []
  for i in 1..int 
    retary << ary
  end
  retary
end

2.step(8){|i|
  ans += @ary357.product(*retarys(i,@ary357)).map {|aa| aa.inject(:+).to_i}.count{|num| num < n}
}
puts ans