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
import bisect
import itertools
import sys
mod=10**9 +7
'''fact=[1]*1001
ifact=[1]*1001
for i in range(1,1001):
    fact[i]=((fact[i-1])*i)%mod
    ifact[i]=((ifact[i-1])*pow(i,mod-2,mod))%mod
def ncr(n,r):
    return (((fact[n]*ifact[n-r])%mod)*ifact[r])%mod
def npr(n,r):
    return (((fact[n]*ifact[n-r])%mod))
    '''


def mindiff(a):
    b=a[:]
    b.sort()
    m=10000000000
    for i in range(len(b)-1):
        if b[i+1]-b[i]<m:
            m=b[i+1]-b[i]
    return m
    
def lcm(a,b):
    return a*b//math.gcd(a,b)

    
def merge(a,b):
    i=0;j=0
    c=0
    ans=[]
    while i<len(a) and j<len(b):
        if a[i]<b[j]:
            ans.append(a[i])
            i+=1
        else:
            ans.append(b[j])
            c+=len(a)-i
            j+=1
    ans+=a[i:]
    ans+=b[j:]
    return ans,c
def mergesort(a):
    if len(a)==1:
        return a,0
    mid=len(a)//2   
    left,left_inversion=mergesort(a[:mid])
    right,right_inversion=mergesort(a[mid:])
    m,c=merge(left,right)
    c+=(left_inversion+right_inversion)
    return m,c
    
def is_prime(num):
    if num == 2: return True
    if num == 3: return True
    if num%2 == 0: return False
    if num%3 == 0: return False
    t = 5
    a = 2
    while t <= int(math.sqrt(num)):
        if num%t == 0: return False
        t += a
        a = 6 - a
    return True
    
  
def ceil(a,b):
    if a%b==0:
        return a//b
    else:
        return (a//b + 1)

def binsearch(arr,b,low,high):
    if low==high:
        return low
    if arr[math.ceil((low+high)/2)]<b:
        return binsearch(arr,b,low,math.ceil((low+high)/2) -1 )
    else:
        return binsearch(arr,b,math.ceil((low+high)/2),high)
def ncr1(n,r):
    s=1
    for i in range(min(n-r,r)):
        s*=(n-i)
        s%=mod
        s*=pow(i+1,mod-2,mod)
        s%=mod
    return s
    
def calc(n,m,r):
    s=0
    for i in range(0,r+1,2):
        s+=ncr1(n,i)*ncr1(m,i)
        s%=mod
    return s    
        
        

#/////////////////////////////////////////////////////////////////////////////////////////////////
n=int(input())
i=2
if is_prime(n):
    exit(print(1))
while(i<=int(n**0.5)+1):
    if n%i==0:
        break
    i+=1
n-=i    
exit(print(n//2 +1))

    