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.
import math

def erathostonesSieve(n):
    size = int(math.sqrt(n))+1
    sieve = []
    for i in range(2, int(math.sqrt(n))):
        if n % i == 0: sieve.append(i)
    return sieve

def factorNumber(n, sieve):
    size = int(math.sqrt(n))+1
    for i in range(2, size):
        if n % i == 0 and (i in sieve):
            return i
    return n

n = int(input())

subtractions = 0
sieve = erathostonesSieve(n)

while n > 0:
    d = factorNumber(n, sieve)
    n -= d
    subtractions += 1

print(subtractions)
 	 	 	   	  	 			 	   	 	  				