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.
'''
OPTIMIZED
Notice that once the number becomes even,
it never stops being even as subtracting 2 doesn't change parity.
Thus, the task is to find the smallest divisor, subtract it and print 1+(n−d)/2.
'''


x = int(input())
for a in range(2, 10**6):
    if (x % a == 0):
        print(1 + (x - a) // 2)
        exit()
print(1)


'''
if smallest prime is 2.it means n is even and again and again it will be divided by 2

if n is odd then smallest prime would not be 2,,so some odd number...and   (n-odd smallest prime)
would be even, so we would now subtract 2 again and again..

LONG THING CUT SHORT, AFTER FIRST STEP N WILL BECOME EVEN
AND WE JUST NEEED TO DO ONE STEP AND THEN DIVIDE BY 2


'''