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
import sys
from collections import defaultdict

N = input()
F = defaultdict(int)

def factorization(n):
    f=defaultdict(int)
    a = n
    for k in range(2, n+1):
        while a%k==0:
            f[k]+=1
            a/=k
    return f


def solve():
    A = F.values()
    B = F.keys()
    # (24, 2)
    n = len(A)
    s1 = set()
    for i in range(n):
        a = A[i]
        if a >= 24:
            for j in range(n):
                if i == j: continue
                if A[j] >= 2:
                    a = B[i]
                    b = B[j]
                    s1.add(a**24 + b**2)
    # (14, 4)
    n = len(A)
    s2 = set()
    for i in range(n):
        a = A[i]
        if a >= 14:
            for j in range(n):
                if i == j: continue
                if A[j] >= 4:
                    a = B[i]
                    b = B[j]
                    s2.add(a**14 + b**4)

    # (4, 4, 2)
    n = len(A)
    s3 = set()
    for i in range(n):
        a = A[i]
        if a >= 4:
            for j in range(n):
                if i == j: continue
                if A[j] >= 4:
                    for k in range(n):
                        if k == i or k == j: continue
                        if A[k] >= 2:
                            a = B[i]
                            b = B[j]
                            c = B[k]
                            s3.add(a**4 + b**4+c**2)

    # (74)    
    n = len(A)
    s4 = set()
    for i in range(n):
        a = A[i]
        if a >= 74:
            s4.add(a**74)                    

    return len(s1) + len(s2) + len(s3) + len(s4)

    




for n in range(2, N+1):
    f = factorization(n)
    for k, v in f.iteritems():
        F[k] += v

print(solve())
