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.

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!.

Examples

Input

9


Output

0


Input

10


Output

1


Input

100


Output

543
from math import sqrt
from bisect import bisect_left as bi
n = int(input())
x,z = [0]*98,[]
for i in range(2,n+1):
	m = i
	while 1:
		f = 0
		for j in range(2,int(sqrt(m))+1):
			if m%j==0:
				x[j]+=1
				f = 1
				break
		if f: m//=j
		else:
			x[m]+=1
			break
for i in x[::-1]:
	if i>1: z.append(i)
p = len(z)
print(p-bi(z,74)+max((p-bi(z,24))*(p-bi(z,2)-1),0)+max((p-bi(z,14))*(p-bi(z,4)-1),0)+max((p-bi(z,4))*(p-bi(z,4)-1)*(p-bi(z,2)-2)//2,0))