Problem:
Score : 500 points

Problem Statement

For an integer n not less than 0, let us define f(n) as follows:

f(n) = 1 (if n < 2)

f(n) = n f(n-2) (if n \geq 2)

Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).

Constraints

0 \leq N \leq 10^{18}

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of trailing zeros in the decimal notation of f(N).

Sample Input 1

12

Sample Output 1

1

f(12) = 12 × 10 × 8 × 6 × 4 × 2 = 46080, which has one trailing zero.

Sample Input 2

5

Sample Output 2

0

f(5) = 5 × 3 × 1 = 15, which has no trailing zeros.

Sample Input 3

1000000000000000000

Sample Output 3

124999999999999995

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
typedef long long ll ;
using namespace std ;
ll t=0,e=0,o=0,ans=0;
void digit(ll n)
{
    ll k=5,x;
    while(k<=n)
    {
        ll p ;
        p=(ll)(n/k);
        t+=p;
        e+=(ll)(p/2);
        k*=5;
    }
}
int main()
{
    ll n ;
    cin >> n ;
    digit(n);
    o=t-e;
    if(n%2==0)
    {
        cout << e << endl;
    }
    else
    {
        cout << "0" << endl;
    }
    return 0;
}