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 primefactors(n):
    i=2
    blanck=[]
    while i**2<=n :
        if n%i==0:
            blanck.append(i)
            while n%i==0:
                n//=i
        else:
            i+=1
    if n>1:
        blanck.append(n)
        n=1
    return blanck
def brute(n):
    cnt=0
    while n:
        x=primefactors(n)
        n-=x[0]
        cnt+=1
    return cnt
def f(n):
    t=primefactors(n)[0]
    if t==n:
        return 1
    else:
        return n//2
n=int(input())
# for i in range(2,1000):
#     if f(i)!=brute(i):
#         print(f(i),brute(i),i,i//2,len(primefactors(i)))
print(f(n))