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 Go (1.6) (judge verdict: Accepted)

package main

import "fmt"

func pf(n int) []int {
	pf := make([]int, 101)
	pf[1]++
	for i := 2; i <= n; i++ {
		x := i
		for j := 2; j <= i; j++ {
			for x%j == 0 {
				pf[j]++
				x /= j
			}
		}
	}
	return pf
}

func num(e []int, x int) int {
	ct := 0
	for _, v := range e {
		if v >= x {
			ct++
		}
	}
	return ct
}

func main() {
	var n int
	fmt.Scan(&n)
	e := pf(n)
	ans := 0
	ans += num(e, 74)
	ans += num(e, 24) * (num(e, 2) - 1)
	ans += num(e, 14) * (num(e, 4) - 1)
	ans += num(e, 4) * (num(e, 4) - 1) * (num(e, 2) - 2) / 2
	fmt.Println(ans)
}