You are given an integer number n. The following algorithm is applied to it:

  1. if n = 0, then end algorithm; 
  2. find the smallest prime divisor d of n; 
  3. subtract d from n and go to step 1. 



Determine the number of subtrations the algorithm will make.

Input

The only line contains a single integer n (2 ≤ n ≤ 10^{10}).

Output

Print a single integer — the number of subtractions the algorithm will make.

Examples

Input


5


Output


1


Input


4


Output


2

Note

In the first example 5 is the smallest prime divisor, thus it gets subtracted right away to make a 0.

In the second example 2 is the smallest prime divisor at both steps.
def factorize(n):  # o(sqr(n))
    c, ans = 1, []
    while (c * c < n):
        if n % c == 0:
            ans.extend([c, n // c])
        c += 1

    if c * c == n:
        ans.append(c)
    return sorted(ans)


def prime(n):
    if n in [2, 3]:
        return True
    if n < 2 or n % 2 == 0 or n % 3 == 0:
        return False
    i = 5
    while (i * i <= n):
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6
    return True


n = int(input())
fac = factorize(n)
for j in fac:
    if prime(j):
        print(n // j)
        exit()
