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: Wrong Answer)

{
	75=3*25
		=3*5*5
		=5*15
	a^2 b^24
	a^2 b^4 c^4
	a^4 b^14
}
var
	i,j,k,ans,n:longint;
	s:array[0..110] of longint;
	f:array[0..110,0..110,0..110] of boolean;
procedure prime(x:longint);
begin
	j:=2;
	while x>=j do
	begin
		while x mod j=0 do
		begin
			inc(s[j]);
			x:=x div j;
		end;
		inc(j);
	end;
end;
begin
	read(n);
	fillchar(s,sizeof(s),0);
	for i:=2 to n do prime(i);
	for i:=2 to n do writeln('s[',i,']=',s[i]);
	fillchar(f,sizeof(f),true);
	ans:=0;
	for i:=2 to n do
		for j:=2 to n do
			if i<>j then
			begin
				if (s[i]>=2) and (s[j]>=24) then inc(ans);
				if (s[i]>=4) and (s[j]>=14) then inc(ans);
				if (s[i]>=2) and (s[j]>=4) then 
					for k:=2 to n do
						if (i<>k) and (j<>k) then
							if s[k]>=4 then 
								if f[i,j,k] then
								begin 
									inc(ans);
									f[i,j,k]:=false;
									f[i,k,j]:=false;
								end;
			end;
	for i:=1 to n do
		if s[i]>=74 then inc(ans);
	writeln(ans);
end.