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
n = int(raw_input())

def prime_div(n):
    arr = []
    f = 1
    while f:
        state = 0
        for i in range(2,int(n**2)+1):
            if n % i == 0:
                arr.append(i)
                n /= i
                state = 1
                break
        if state == 0:
            f = 0
    return arr
arr = []
for i in range(n):
    arr += prime_div(i+1)
arr.sort()
num_set = set(arr)
# print num_set
counter = []
counter_num = [0,0,0,0,0]
num_lst = [2, 4, 14, 24, 74]
for num in num_set:
    count = arr.count(num)
    counter.append(count)
    for i in range(5):
        if count >= num_lst[i]:
            counter_num[i]+=1

# print counter_num
ans = 0
# 74
ans += counter_num[4]
# 24 2
if counter_num[3] > 0:
    ans += counter_num[3]*(counter_num[0]-counter_num[3])+counter_num[3]*(counter_num[3]-1)
# 14 4
if counter_num[2] > 0:
    ans += counter_num[2]*(counter_num[1]-counter_num[2])+counter_num[2]*(counter_num[2]-1)
if counter_num[1] > 1:
    ans += counter_num[1]*(counter_num[1]-1)/2*(counter_num[0]-counter_num[1])
if counter_num[1] > 2:
    ans += counter_num[1]*(counter_num[1]-1)/2*(counter_num[1]-2)
print ans
