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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

var
	n,i,ans,p,j,a,cnt,k:Longint;
	dp:array[0..100,1..75]of Longint;
function isp(a:Longint):Boolean;
var i:Longint;
begin
	i:=2;
	while i*i<=a do begin
		if a mod i=0 then begin
			isp:=False;
			exit;
		end;
		inc(i);
	end;
	isp:=True;
end;
begin
	read(n);
	dp[0,1]:=1;
	p:=0;
	for i:=2 to n do begin
		if not isp(i)then continue;
		cnt:=0;
		for j:=1 to n do begin
			a:=j;
			while a mod i=0 do begin
				a:=a div i;
				inc(cnt);
			end;
		end;
		for j:=0 to cnt do begin
			if not((j=0)or(j=2)or(j=4)or(j=14)or(j=24)or(j=74))then continue;
			for k:=1 to 75 div (j+1)do inc(dp[i,k*(j+1)],dp[p,k]);
		end;
		p:=i;
	end;
	writeln(dp[p,75]);
end.