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 LuaJIT (2.0.4) (judge verdict: Accepted)

local mfl, mce, msq = math.floor, math.ceil, math.sqrt
local n = io.read("*n")

local function getprimes(x)
  local primes = {}
  local allnums = {}
  for i = 1, x do allnums[i] = true end
  for i = 2, x do
    if(allnums[i]) then
      table.insert(primes, i)
      local lim = mfl(x / i)
      for j = 2, lim do
        allnums[j * i] = false
      end
    end
  end
  return primes
end

local primes = getprimes(47)
local t = {}
for i = 1, 47 do
  t[i] = 0
end
for i = 1, #primes do
  local p = primes[i]
  local ctot = 0
  local c = mfl(n / p)
  while 0 < c do
    ctot = ctot + c
    c = mfl(c / p)
  end
  t[p] = ctot
end

local ov74, ov24, ov14, ov4, ov2 = 0, 0, 0, 0, 0
for i = 1, 47 do
  if 74 <= t[i] then ov74 = ov74 + 1 end
  if 24 <= t[i] then ov24 = ov24 + 1 end
  if 14 <= t[i] then ov14 = ov14 + 1 end
  if 4 <= t[i] then ov4 = ov4 + 1 end
  if 2 <= t[i] then ov2 = ov2 + 1 end
end
local ret = ov74 + ov24 * (ov2 - 1) + ov14 * (ov4 - 1) + mfl(ov4 * (ov4 - 1) / 2) * (ov2 - 2)
print(ret)