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
#!/usr/bin/env python

from collections import deque
import itertools as it
import sys
import math

sys.setrecursionlimit(1000000)

INF = 10 ** 18
MOD = 10 ** 9 + 7

N = input()
cnt = [1] * 110
for num in range(1, N + 1):
    for i in range(2, 100):
        while num % i == 0:
            cnt[i] += 1
            num /= i

ans = 0
for i in range(N + 1):
    for j in range(N + 1):
        for k in range(N + 1):
            if i == j or j == k or k == i:
                continue
            if cnt[i] >= 3 and cnt[j] >= 5 and cnt[k] >= 5:
                ans += 1

ans /= 2

for i in range(N + 1):
    for j in range(N + 1):
        if i == j:
            continue
        if cnt[i] >= 5 and cnt[j] >= 15:
            ans += 1
        if cnt[i] >= 3 and cnt[j] >= 25:
            ans += 1
    if cnt[i] >= 75:
        ans += 1

print ans