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 list_primes_upto(upto :int):
    uptop1 = upto+1
    uptom1 = upto-1
    pcl = [True for x in range(uptop1)]
    lprimes = []
    if upto >= 2: lprimes.append(2)
    if upto >= 3: lprimes.append(3)
    i = 5
    while i < uptom1:
        ix2 = i*2
        if pcl[i]:
            lprimes.append(i)
            for k in range(i*i, uptop1, ix2):
                pcl[k] = False
        ip2 = i+2
        ip2x2 = ip2*2
        if pcl[ip2]:
            lprimes.append(ip2)
            for k in range(ip2*ip2, uptop1, ip2x2):
                pcl[k] = False
        i += 6
    if i <= upto:
        ix2 = i*2
        if pcl[i]:
            lprimes.append(i)
            for k in range(i*i, uptop1, ix2):
                pcl[k] = False
    return lprimes


n = int(input())
lp = list_primes_upto(int(math.sqrt(n)) + 1)
subs = 0
pi = 0
while n != 0 and pi < len(lp):
    if n % lp[pi] == 0:
        if pi == 0:
            subs += n // 2
            n = 0
        else:
            n -= lp[pi]
            subs += 1
            pi = 0
    else:
        pi += 1
if n != 0:
    subs += 1
print(subs)
