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 configparser
import sys
input = sys.stdin.readline


def ceil_div(x, y):
    return -(-x // y)

def main():
    n = int(input())

    if n % 2 == 0:
        print(n // 2)
        return


    import math

    max_n = int(math.sqrt(n))+5
    prime = [True for i in range(max_n)]

    for i in range(2, max_n):
        if prime[i]:
            for j in range(i+i, max_n , i):
                prime[j] = False


    find = -1
    for i in range(3, max_n):
        if i*i >n:
            break
        if n%i==0 and prime[i]:
            find = i
            break

    if find == -1:
        print(1)
    else:
        print(1 + (n-find)//2)

if __name__ == '__main__':
    main()
